What is the general solution for cos^{22}(x)-2sin^2(x)-1=0 ?

The general solution for cos^{22}(x)-2sin^2(x)-1=0 is x=2pin,x=pi+2pin

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cos^{22}(x)-2sin^2(x)-1=0

Solution

cos^{22} (x) - 2 sin^{2} (x) - 1 = 0

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{22} (x) - 2 sin^{2} (x) - 1 = 0

Rewrite using trig identities

- 1 + cos^{22} (x) - 2 sin^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 1 + cos^{22} (x) - 2 (1 - cos^{2} (x))

Simplify

- 1 + cos^{22} (x) - 2 (1 - cos^{2} (x)) : cos^{22} (x) + 2 cos^{2} (x) - 3

- 1 + cos^{22} (x) - 2 (1 - cos^{2} (x))

Expand

- 2 (1 - cos^{2} (x)) : - 2 + 2 cos^{2} (x)

- 2 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (x)

= - 2 \cdot 1 - (- 2) cos^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 cos^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 cos^{2} (x)

= - 1 + cos^{22} (x) - 2 + 2 cos^{2} (x)

Simplify

- 1 + cos^{22} (x) - 2 + 2 cos^{2} (x) : cos^{22} (x) + 2 cos^{2} (x) - 3

- 1 + cos^{22} (x) - 2 + 2 cos^{2} (x)

Group like terms

= cos^{22} (x) + 2 cos^{2} (x) - 1 - 2

Subtract the numbers:

- 1 - 2 = - 3

= cos^{22} (x) + 2 cos^{2} (x) - 3

= cos^{22} (x) + 2 cos^{2} (x) - 3

= cos^{22} (x) + 2 cos^{2} (x) - 3

- 3 + cos^{22} (x) + 2 cos^{2} (x) = 0

Solve by substitution

- 3 + cos^{22} (x) + 2 cos^{2} (x) = 0

Let:

cos (x) = u

- 3 + u^{22} + 2 u^{2} = 0

- 3 + u^{22} + 2 u^{2} = 0 : u = 1, u = - 1

- 3 + u^{22} + 2 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{22} + 2 u^{2} - 3 = 0

Rewrite the equation with

v = u^{2}

and

v^{11} = u^{22}

v^{11} + 2 v - 3 = 0

Solve

v^{11} + 2 v - 3 = 0 : v = 1

v^{11} + 2 v - 3 = 0

Factor

v^{11} + 2 v - 3 : (v - 1) (v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3)

v^{11} + 2 v - 3

Use the rational root theorem

a_{0} = 3, a_{n} = 1

The dividers of

a_{0} : 1, 3,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1 , 3}{1}

\frac{1}{1}

is a root of the expression, so factor out

v - 1

= (v - 1) \frac{v ^{11} + 2 v - 3}{v - 1}

\frac{v ^{11} + 2 v - 3}{v - 1} = v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3

\frac{v ^{11} + 2 v - 3}{v - 1}

Divide

\frac{v ^{11} + 2 v - 3}{v - 1} : \frac{v ^{11} + 2 v - 3}{v - 1} = v^{10} + \frac{v ^{10} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{11} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{11}}{v} = v^{10}

Quotient = v^{10}

Multiply

v - 1

v^{10} : v^{11} - v^{10}

Subtract

v^{11} - v^{10}

from

v^{11} + 2 v - 3

to get new remainder

Remainder = v^{10} + 2 v - 3

Therefore

\frac{v ^{11} + 2 v - 3}{v - 1} = v^{10} + \frac{v ^{10} + 2 v - 3}{v - 1}

= v^{10} + \frac{v ^{10} + 2 v - 3}{v - 1}

Divide

\frac{v ^{10} + 2 v - 3}{v - 1} : \frac{v ^{10} + 2 v - 3}{v - 1} = v^{9} + \frac{v ^{9} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{10} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{10}}{v} = v^{9}

Quotient = v^{9}

Multiply

v - 1

v^{9} : v^{10} - v^{9}

Subtract

v^{10} - v^{9}

from

v^{10} + 2 v - 3

to get new remainder

Remainder = v^{9} + 2 v - 3

Therefore

\frac{v ^{10} + 2 v - 3}{v - 1} = v^{9} + \frac{v ^{9} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + \frac{v ^{9} + 2 v - 3}{v - 1}

Divide

\frac{v ^{9} + 2 v - 3}{v - 1} : \frac{v ^{9} + 2 v - 3}{v - 1} = v^{8} + \frac{v ^{8} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{9} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{9}}{v} = v^{8}

Quotient = v^{8}

Multiply

v - 1

v^{8} : v^{9} - v^{8}

Subtract

v^{9} - v^{8}

from

v^{9} + 2 v - 3

to get new remainder

Remainder = v^{8} + 2 v - 3

Therefore

\frac{v ^{9} + 2 v - 3}{v - 1} = v^{8} + \frac{v ^{8} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + \frac{v ^{8} + 2 v - 3}{v - 1}

Divide

\frac{v ^{8} + 2 v - 3}{v - 1} : \frac{v ^{8} + 2 v - 3}{v - 1} = v^{7} + \frac{v ^{7} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{8} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{8}}{v} = v^{7}

Quotient = v^{7}

Multiply

v - 1

v^{7} : v^{8} - v^{7}

Subtract

v^{8} - v^{7}

from

v^{8} + 2 v - 3

to get new remainder

Remainder = v^{7} + 2 v - 3

Therefore

\frac{v ^{8} + 2 v - 3}{v - 1} = v^{7} + \frac{v ^{7} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + \frac{v ^{7} + 2 v - 3}{v - 1}

Divide

\frac{v ^{7} + 2 v - 3}{v - 1} : \frac{v ^{7} + 2 v - 3}{v - 1} = v^{6} + \frac{v ^{6} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{7} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{7}}{v} = v^{6}

Quotient = v^{6}

Multiply

v - 1

v^{6} : v^{7} - v^{6}

Subtract

v^{7} - v^{6}

from

v^{7} + 2 v - 3

to get new remainder

Remainder = v^{6} + 2 v - 3

Therefore

\frac{v ^{7} + 2 v - 3}{v - 1} = v^{6} + \frac{v ^{6} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + \frac{v ^{6} + 2 v - 3}{v - 1}

Divide

\frac{v ^{6} + 2 v - 3}{v - 1} : \frac{v ^{6} + 2 v - 3}{v - 1} = v^{5} + \frac{v ^{5} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{6} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{6}}{v} = v^{5}

Quotient = v^{5}

Multiply

v - 1

v^{5} : v^{6} - v^{5}

Subtract

v^{6} - v^{5}

from

v^{6} + 2 v - 3

to get new remainder

Remainder = v^{5} + 2 v - 3

Therefore

\frac{v ^{6} + 2 v - 3}{v - 1} = v^{5} + \frac{v ^{5} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + \frac{v ^{5} + 2 v - 3}{v - 1}

Divide

\frac{v ^{5} + 2 v - 3}{v - 1} : \frac{v ^{5} + 2 v - 3}{v - 1} = v^{4} + \frac{v ^{4} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{5} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{5}}{v} = v^{4}

Quotient = v^{4}

Multiply

v - 1

v^{4} : v^{5} - v^{4}

Subtract

v^{5} - v^{4}

from

v^{5} + 2 v - 3

to get new remainder

Remainder = v^{4} + 2 v - 3

Therefore

\frac{v ^{5} + 2 v - 3}{v - 1} = v^{4} + \frac{v ^{4} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + \frac{v ^{4} + 2 v - 3}{v - 1}

Divide

\frac{v ^{4} + 2 v - 3}{v - 1} : \frac{v ^{4} + 2 v - 3}{v - 1} = v^{3} + \frac{v ^{3} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{4} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{4}}{v} = v^{3}

Quotient = v^{3}

Multiply

v - 1

v^{3} : v^{4} - v^{3}

Subtract

v^{4} - v^{3}

from

v^{4} + 2 v - 3

to get new remainder

Remainder = v^{3} + 2 v - 3

Therefore

\frac{v ^{4} + 2 v - 3}{v - 1} = v^{3} + \frac{v ^{3} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + \frac{v ^{3} + 2 v - 3}{v - 1}

Divide

\frac{v ^{3} + 2 v - 3}{v - 1} : \frac{v ^{3} + 2 v - 3}{v - 1} = v^{2} + \frac{v ^{2} + 2 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{3} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{3}}{v} = v^{2}

Quotient = v^{2}

Multiply

v - 1

v^{2} : v^{3} - v^{2}

Subtract

v^{3} - v^{2}

from

v^{3} + 2 v - 3

to get new remainder

Remainder = v^{2} + 2 v - 3

Therefore

\frac{v ^{3} + 2 v - 3}{v - 1} = v^{2} + \frac{v ^{2} + 2 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + \frac{v ^{2} + 2 v - 3}{v - 1}

Divide

\frac{v ^{2} + 2 v - 3}{v - 1} : \frac{v ^{2} + 2 v - 3}{v - 1} = v + \frac{3 v - 3}{v - 1}

Divide the leading coefficients of the numerator

v^{2} + 2 v - 3

and the divisor

v - 1 : \frac{v ^{2}}{v} = v

Quotient = v

Multiply

v - 1

v : v^{2} - v

Subtract

v^{2} - v

from

v^{2} + 2 v - 3

to get new remainder

Remainder = 3 v - 3

Therefore

\frac{v ^{2} + 2 v - 3}{v - 1} = v + \frac{3 v - 3}{v - 1}

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + \frac{3 v - 3}{v - 1}

Divide

\frac{3 v - 3}{v - 1} : \frac{3 v - 3}{v - 1} = 3

Divide the leading coefficients of the numerator

3 v - 3

and the divisor

v - 1 : \frac{3 v}{v} = 3

Quotient = 3

Multiply

v - 1

3 : 3 v - 3

Subtract

3 v - 3

from

3 v - 3

to get new remainder

Remainder = 0

Therefore

\frac{3 v - 3}{v - 1} = 3

= v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3

= (v - 1) (v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3)

(v - 1) (v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v - 1 = 0 or v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3 = 0

Solve

v - 1 = 0 : v = 1

v - 1 = 0

Move

1

to the right side

v - 1 = 0

Add

1

to both sides

v - 1 + 1 = 0 + 1

Simplify

v = 1

v = 1

Solve

v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3 = 0 :

No Solution for

v \in R

v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3 = 0

Find one solution for

v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3 = 0

using Newton-Raphson:

No Solution for

v \in R

v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3 = 0

Newton-Raphson Approximation Definition

f (v) = v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3

Find

f^{^{'}} (v) : 10 v^{9} + 9 v^{8} + 8 v^{7} + 7 v^{6} + 6 v^{5} + 5 v^{4} + 4 v^{3} + 3 v^{2} + 2 v + 1

\frac{d}{d v} (v^{10} + v^{9} + v^{8} + v^{7} + v^{6} + v^{5} + v^{4} + v^{3} + v^{2} + v + 3)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d v} (v^{10}) + \frac{d}{d v} (v^{9}) + \frac{d}{d v} (v^{8}) + \frac{d}{d v} (v^{7}) + \frac{d}{d v} (v^{6}) + \frac{d}{d v} (v^{5}) + \frac{d}{d v} (v^{4}) + \frac{d}{d v} (v^{3}) + \frac{d}{d v} (v^{2}) + \frac{d v}{d v} + \frac{d}{d v} (3)

\frac{d}{d v} (v^{10}) = 10 v^{9}

\frac{d}{d v} (v^{10})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 10 v^{10 - 1}

Simplify

= 10 v^{9}

\frac{d}{d v} (v^{9}) = 9 v^{8}

\frac{d}{d v} (v^{9})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 9 v^{9 - 1}

Simplify

= 9 v^{8}

\frac{d}{d v} (v^{8}) = 8 v^{7}

\frac{d}{d v} (v^{8})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 8 v^{8 - 1}

Simplify

= 8 v^{7}

\frac{d}{d v} (v^{7}) = 7 v^{6}

\frac{d}{d v} (v^{7})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 7 v^{7 - 1}

Simplify

= 7 v^{6}

\frac{d}{d v} (v^{6}) = 6 v^{5}

\frac{d}{d v} (v^{6})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 v^{6 - 1}

Simplify

= 6 v^{5}

\frac{d}{d v} (v^{5}) = 5 v^{4}

\frac{d}{d v} (v^{5})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 v^{5 - 1}

Simplify

= 5 v^{4}

\frac{d}{d v} (v^{4}) = 4 v^{3}

\frac{d}{d v} (v^{4})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 v^{4 - 1}

Simplify

= 4 v^{3}

\frac{d}{d v} (v^{3}) = 3 v^{2}

\frac{d}{d v} (v^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 v^{3 - 1}

Simplify

= 3 v^{2}

\frac{d}{d v} (v^{2}) = 2 v

\frac{d}{d v} (v^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 v^{2 - 1}

Simplify

= 2 v

\frac{d v}{d v} = 1

\frac{d v}{d v}

Apply the common derivative:

\frac{d v}{d v} = 1

= 1

\frac{d}{d v} (3) = 0

\frac{d}{d v} (3)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 10 v^{9} + 9 v^{8} + 8 v^{7} + 7 v^{6} + 6 v^{5} + 5 v^{4} + 4 v^{3} + 3 v^{2} + 2 v + 1 + 0

Simplify

= 10 v^{9} + 9 v^{8} + 8 v^{7} + 7 v^{6} + 6 v^{5} + 5 v^{4} + 4 v^{3} + 3 v^{2} + 2 v + 1

Let

v_{0} = - 3

Compute

v_{n + 1}

until

Δ v_{n + 1} < 0.000001

v_{1} = - 2.70730 \dots : Δ v_{1} = 0.29269 \dots

f (v_{0}) = (- 3)^{10} + (- 3)^{9} + (- 3)^{8} + (- 3)^{7} + (- 3)^{6} + (- 3)^{5} + (- 3)^{4} + (- 3)^{3} + (- 3)^{2} + (- 3) + 3 = 44289

f^{^{'}} (v_{0}) = 10 (- 3)^{9} + 9 (- 3)^{8} + 8 (- 3)^{7} + 7 (- 3)^{6} + 6 (- 3)^{5} + 5 (- 3)^{4} + 4 (- 3)^{3} + 3 (- 3)^{2} + 2 (- 3) + 1 = - 151313

v_{1} = - 2.70730 \dots

Δ v_{1} = ∣ - 2.70730 \dots - (- 3) ∣ = 0.29269 \dots

Δ v_{1} = 0.29269 \dots

v_{2} = - 2.44364 \dots : Δ v_{2} = 0.26365 \dots

f (v_{1}) = (- 2.70730 \dots)^{10} + (- 2.70730 \dots)^{9} + (- 2.70730 \dots)^{8} + (- 2.70730 \dots)^{7} + (- 2.70730 \dots)^{6} + (- 2.70730 \dots)^{5} + (- 2.70730 \dots)^{4} + (- 2.70730 \dots)^{3} + (- 2.70730 \dots)^{2} + (- 2.70730 \dots) + 3 = 15449.33265 \dots

f^{^{'}} (v_{1}) = 10 (- 2.70730 \dots)^{9} + 9 (- 2.70730 \dots)^{8} + 8 (- 2.70730 \dots)^{7} + 7 (- 2.70730 \dots)^{6} + 6 (- 2.70730 \dots)^{5} + 5 (- 2.70730 \dots)^{4} + 4 (- 2.70730 \dots)^{3} + 3 (- 2.70730 \dots)^{2} + 2 (- 2.70730 \dots) + 1 = - 58596.00670 \dots

v_{2} = - 2.44364 \dots

Δ v_{2} = ∣ - 2.44364 \dots - (- 2.70730 \dots) ∣ = 0.26365 \dots

Δ v_{2} = 0.26365 \dots

v_{3} = - 2.20607 \dots : Δ v_{3} = 0.23757 \dots

f (v_{2}) = (- 2.44364 \dots)^{10} + (- 2.44364 \dots)^{9} + (- 2.44364 \dots)^{8} + (- 2.44364 \dots)^{7} + (- 2.44364 \dots)^{6} + (- 2.44364 \dots)^{5} + (- 2.44364 \dots)^{4} + (- 2.44364 \dots)^{3} + (- 2.44364 \dots)^{2} + (- 2.44364 \dots) + 3 = 5389.92791 \dots

f^{^{'}} (v_{2}) = 10 (- 2.44364 \dots)^{9} + 9 (- 2.44364 \dots)^{8} + 8 (- 2.44364 \dots)^{7} + 7 (- 2.44364 \dots)^{6} + 6 (- 2.44364 \dots)^{5} + 5 (- 2.44364 \dots)^{4} + 4 (- 2.44364 \dots)^{3} + 3 (- 2.44364 \dots)^{2} + 2 (- 2.44364 \dots) + 1 = - 22687.71319 \dots

v_{3} = - 2.20607 \dots

Δ v_{3} = ∣ - 2.20607 \dots - (- 2.44364 \dots) ∣ = 0.23757 \dots

Δ v_{3} = 0.23757 \dots

v_{4} = - 1.99187 \dots : Δ v_{4} = 0.21420 \dots

f (v_{3}) = (- 2.20607 \dots)^{10} + (- 2.20607 \dots)^{9} + (- 2.20607 \dots)^{8} + (- 2.20607 \dots)^{7} + (- 2.20607 \dots)^{6} + (- 2.20607 \dots)^{5} + (- 2.20607 \dots)^{4} + (- 2.20607 \dots)^{3} + (- 2.20607 \dots)^{2} + (- 2.20607 \dots) + 3 = 1880.96171 \dots

f^{^{'}} (v_{3}) = 10 (- 2.20607 \dots)^{9} + 9 (- 2.20607 \dots)^{8} + 8 (- 2.20607 \dots)^{7} + 7 (- 2.20607 \dots)^{6} + 6 (- 2.20607 \dots)^{5} + 5 (- 2.20607 \dots)^{4} + 4 (- 2.20607 \dots)^{3} + 3 (- 2.20607 \dots)^{2} + 2 (- 2.20607 \dots) + 1 = - 8781.32634 \dots

v_{4} = - 1.99187 \dots

Δ v_{4} = ∣ - 1.99187 \dots - (- 2.20607 \dots) ∣ = 0.21420 \dots

Δ v_{4} = 0.21420 \dots

v_{5} = - 1.79843 \dots : Δ v_{5} = 0.19343 \dots

f (v_{4}) = (- 1.99187 \dots)^{10} + (- 1.99187 \dots)^{9} + (- 1.99187 \dots)^{8} + (- 1.99187 \dots)^{7} + (- 1.99187 \dots)^{6} + (- 1.99187 \dots)^{5} + (- 1.99187 \dots)^{4} + (- 1.99187 \dots)^{3} + (- 1.99187 \dots)^{2} + (- 1.99187 \dots) + 3 = 656.87267 \dots

f^{^{'}} (v_{4}) = 10 (- 1.99187 \dots)^{9} + 9 (- 1.99187 \dots)^{8} + 8 (- 1.99187 \dots)^{7} + 7 (- 1.99187 \dots)^{6} + 6 (- 1.99187 \dots)^{5} + 5 (- 1.99187 \dots)^{4} + 4 (- 1.99187 \dots)^{3} + 3 (- 1.99187 \dots)^{2} + 2 (- 1.99187 \dots) + 1 = - 3395.76549 \dots

v_{5} = - 1.79843 \dots

Δ v_{5} = ∣ - 1.79843 \dots - (- 1.99187 \dots) ∣ = 0.19343 \dots

Δ v_{5} = 0.19343 \dots

v_{6} = - 1.62297 \dots : Δ v_{6} = 0.17545 \dots

f (v_{5}) = (- 1.79843 \dots)^{10} + (- 1.79843 \dots)^{9} + (- 1.79843 \dots)^{8} + (- 1.79843 \dots)^{7} + (- 1.79843 \dots)^{6} + (- 1.79843 \dots)^{5} + (- 1.79843 \dots)^{4} + (- 1.79843 \dots)^{3} + (- 1.79843 \dots)^{2} + (- 1.79843 \dots) + 3 = 229.82796 \dots

f^{^{'}} (v_{5}) = 10 (- 1.79843 \dots)^{9} + 9 (- 1.79843 \dots)^{8} + 8 (- 1.79843 \dots)^{7} + 7 (- 1.79843 \dots)^{6} + 6 (- 1.79843 \dots)^{5} + 5 (- 1.79843 \dots)^{4} + 4 (- 1.79843 \dots)^{3} + 3 (- 1.79843 \dots)^{2} + 2 (- 1.79843 \dots) + 1 = - 1309.89577 \dots

v_{6} = - 1.62297 \dots

Δ v_{6} = ∣ - 1.62297 \dots - (- 1.79843 \dots) ∣ = 0.17545 \dots

Δ v_{6} = 0.17545 \dots

v_{7} = - 1.46185 \dots : Δ v_{7} = 0.16112 \dots

f (v_{6}) = (- 1.62297 \dots)^{10} + (- 1.62297 \dots)^{9} + (- 1.62297 \dots)^{8} + (- 1.62297 \dots)^{7} + (- 1.62297 \dots)^{6} + (- 1.62297 \dots)^{5} + (- 1.62297 \dots)^{4} + (- 1.62297 \dots)^{3} + (- 1.62297 \dots)^{2} + (- 1.62297 \dots) + 3 = 80.84111 \dots

f^{^{'}} (v_{6}) = 10 (- 1.62297 \dots)^{9} + 9 (- 1.62297 \dots)^{8} + 8 (- 1.62297 \dots)^{7} + 7 (- 1.62297 \dots)^{6} + 6 (- 1.62297 \dots)^{5} + 5 (- 1.62297 \dots)^{4} + 4 (- 1.62297 \dots)^{3} + 3 (- 1.62297 \dots)^{2} + 2 (- 1.62297 \dots) + 1 = - 501.71642 \dots

v_{7} = - 1.46185 \dots

Δ v_{7} = ∣ - 1.46185 \dots - (- 1.62297 \dots) ∣ = 0.16112 \dots

Δ v_{7} = 0.16112 \dots

v_{8} = - 1.30846 \dots : Δ v_{8} = 0.15338 \dots

f (v_{7}) = (- 1.46185 \dots)^{10} + (- 1.46185 \dots)^{9} + (- 1.46185 \dots)^{8} + (- 1.46185 \dots)^{7} + (- 1.46185 \dots)^{6} + (- 1.46185 \dots)^{5} + (- 1.46185 \dots)^{4} + (- 1.46185 \dots)^{3} + (- 1.46185 \dots)^{2} + (- 1.46185 \dots) + 3 = 28.87105 \dots

f^{^{'}} (v_{7}) = 10 (- 1.46185 \dots)^{9} + 9 (- 1.46185 \dots)^{8} + 8 (- 1.46185 \dots)^{7} + 7 (- 1.46185 \dots)^{6} + 6 (- 1.46185 \dots)^{5} + 5 (- 1.46185 \dots)^{4} + 4 (- 1.46185 \dots)^{3} + 3 (- 1.46185 \dots)^{2} + 2 (- 1.46185 \dots) + 1 = - 188.22544 \dots

v_{8} = - 1.30846 \dots

Δ v_{8} = ∣ - 1.30846 \dots - (- 1.46185 \dots) ∣ = 0.15338 \dots

Δ v_{8} = 0.15338 \dots

v_{9} = - 1.14599 \dots : Δ v_{9} = 0.16247 \dots

f (v_{8}) = (- 1.30846 \dots)^{10} + (- 1.30846 \dots)^{9} + (- 1.30846 \dots)^{8} + (- 1.30846 \dots)^{7} + (- 1.30846 \dots)^{6} + (- 1.30846 \dots)^{5} + (- 1.30846 \dots)^{4} + (- 1.30846 \dots)^{3} + (- 1.30846 \dots)^{2} + (- 1.30846 \dots) + 3 = 10.77112 \dots

f^{^{'}} (v_{8}) = 10 (- 1.30846 \dots)^{9} + 9 (- 1.30846 \dots)^{8} + 8 (- 1.30846 \dots)^{7} + 7 (- 1.30846 \dots)^{6} + 6 (- 1.30846 \dots)^{5} + 5 (- 1.30846 \dots)^{4} + 4 (- 1.30846 \dots)^{3} + 3 (- 1.30846 \dots)^{2} + 2 (- 1.30846 \dots) + 1 = - 66.29578 \dots

v_{9} = - 1.14599 \dots

Δ v_{9} = ∣ - 1.14599 \dots - (- 1.30846 \dots) ∣ = 0.16247 \dots

Δ v_{9} = 0.16247 \dots

v_{10} = - 0.90431 \dots : Δ v_{10} = 0.24167 \dots

f (v_{9}) = (- 1.14599 \dots)^{10} + (- 1.14599 \dots)^{9} + (- 1.14599 \dots)^{8} + (- 1.14599 \dots)^{7} + (- 1.14599 \dots)^{6} + (- 1.14599 \dots)^{5} + (- 1.14599 \dots)^{4} + (- 1.14599 \dots)^{3} + (- 1.14599 \dots)^{2} + (- 1.14599 \dots) + 3 = 4.55228 \dots

f^{^{'}} (v_{9}) = 10 (- 1.14599 \dots)^{9} + 9 (- 1.14599 \dots)^{8} + 8 (- 1.14599 \dots)^{7} + 7 (- 1.14599 \dots)^{6} + 6 (- 1.14599 \dots)^{5} + 5 (- 1.14599 \dots)^{4} + 4 (- 1.14599 \dots)^{3} + 3 (- 1.14599 \dots)^{2} + 2 (- 1.14599 \dots) + 1 = - 18.83635 \dots

v_{10} = - 0.90431 \dots

Δ v_{10} = ∣ - 0.90431 \dots - (- 1.14599 \dots) ∣ = 0.24167 \dots

Δ v_{10} = 0.24167 \dots

v_{11} = 0.64150 \dots : Δ v_{11} = 1.54582 \dots

f (v_{10}) = (- 0.90431 \dots)^{10} + (- 0.90431 \dots)^{9} + (- 0.90431 \dots)^{8} + (- 0.90431 \dots)^{7} + (- 0.90431 \dots)^{6} + (- 0.90431 \dots)^{5} + (- 0.90431 \dots)^{4} + (- 0.90431 \dots)^{3} + (- 0.90431 \dots)^{2} + (- 0.90431 \dots) + 3 = 2.69882 \dots

f^{^{'}} (v_{10}) = 10 (- 0.90431 \dots)^{9} + 9 (- 0.90431 \dots)^{8} + 8 (- 0.90431 \dots)^{7} + 7 (- 0.90431 \dots)^{6} + 6 (- 0.90431 \dots)^{5} + 5 (- 0.90431 \dots)^{4} + 4 (- 0.90431 \dots)^{3} + 3 (- 0.90431 \dots)^{2} + 2 (- 0.90431 \dots) + 1 = - 1.74587 \dots

v_{11} = 0.64150 \dots

Δ v_{11} = ∣ 0.64150 \dots - (- 0.90431 \dots) ∣ = 1.54582 \dots

Δ v_{11} = 1.54582 \dots

v_{12} = - 0.00636 \dots : Δ v_{12} = 0.64787 \dots

f (v_{11}) = 0.64150 \dots^{10} + 0.64150 \dots^{9} + 0.64150 \dots^{8} + 0.64150 \dots^{7} + 0.64150 \dots^{6} + 0.64150 \dots^{5} + 0.64150 \dots^{4} + 0.64150 \dots^{3} + 0.64150 \dots^{2} + 0.64150 \dots + 3 = 4.76834 \dots

f^{^{'}} (v_{11}) = 10 \cdot 0.64150 \dots^{9} + 9 \cdot 0.64150 \dots^{8} + 8 \cdot 0.64150 \dots^{7} + 7 \cdot 0.64150 \dots^{6} + 6 \cdot 0.64150 \dots^{5} + 5 \cdot 0.64150 \dots^{4} + 4 \cdot 0.64150 \dots^{3} + 3 \cdot 0.64150 \dots^{2} + 2 \cdot 0.64150 \dots + 1 = 7.36003 \dots

v_{12} = - 0.00636 \dots

Δ v_{12} = ∣ - 0.00636 \dots - 0.64150 \dots ∣ = 0.64787 \dots

Δ v_{12} = 0.64787 \dots

v_{13} = - 3.03824 \dots : Δ v_{13} = 3.03188 \dots

f (v_{12}) = (- 0.00636 \dots)^{10} + (- 0.00636 \dots)^{9} + (- 0.00636 \dots)^{8} + (- 0.00636 \dots)^{7} + (- 0.00636 \dots)^{6} + (- 0.00636 \dots)^{5} + (- 0.00636 \dots)^{4} + (- 0.00636 \dots)^{3} + (- 0.00636 \dots)^{2} + (- 0.00636 \dots) + 3 = 2.99367 \dots

f^{^{'}} (v_{12}) = 10 (- 0.00636 \dots)^{9} + 9 (- 0.00636 \dots)^{8} + 8 (- 0.00636 \dots)^{7} + 7 (- 0.00636 \dots)^{6} + 6 (- 0.00636 \dots)^{5} + 5 (- 0.00636 \dots)^{4} + 4 (- 0.00636 \dots)^{3} + 3 (- 0.00636 \dots)^{2} + 2 (- 0.00636 \dots) + 1 = 0.98739 \dots

v_{13} = - 3.03824 \dots

Δ v_{13} = ∣ - 3.03824 \dots - (- 0.00636 \dots) ∣ = 3.03188 \dots

Δ v_{13} = 3.03188 \dots

Cannot find solution

The solution is

No Solution for v \in R

The solution is

v = 1

v = 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 1, u = - 1

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

tan(θ)=(3sqrt(5))/2

tan (θ) = \frac{3 5}{2}

sin(x+pi/2)+cos^2(x)=0

sin (x + \frac{π}{2}) + cos^{2} (x) = 0

3sin^2(x)-sin(x)-4=0

3 sin^{2} (x) - sin (x) - 4 = 0

sin(3x)=sin^2(x)

sin (3 x) = sin^{2} (x)

2cot(x)+3=0

2 cot (x) + 3 = 0

Frequently Asked Questions (FAQ)

What is the general solution for cos^{22}(x)-2sin^2(x)-1=0 ?
The general solution for cos^{22}(x)-2sin^2(x)-1=0 is x=2pin,x=pi+2pin