What is the general solution for 4cos(x)=sin^2(x)+1 ?

The general solution for 4cos(x)=sin^2(x)+1 is x=1.10460…+2pin,x=2pi-1.10460…+2pin

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4cos(x)=sin^2(x)+1

Solution

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

Solution

x = 1.10460 \dots + 2 πn, x = 2 π - 1.10460 \dots + 2 πn

Degrees

x = 63.28904 \dots^{\circ} + 36 0^{\circ} n, x = 296.71095 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x) + 1

from both sides

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

Rewrite using trig identities

- 1 - sin^{2} (x) + 4 cos (x)

Use the Pythagorean identity:

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

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

= - 1 - (1 - cos^{2} (x)) + 4 cos (x)

Simplify

- 1 - (1 - cos^{2} (x)) + 4 cos (x) : cos^{2} (x) + 4 cos (x) - 2

- 1 - (1 - cos^{2} (x)) + 4 cos (x)

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

= - 1 - 1 + cos^{2} (x) + 4 cos (x)

Subtract the numbers:

- 1 - 1 = - 2

= cos^{2} (x) + 4 cos (x) - 2

= cos^{2} (x) + 4 cos (x) - 2

- 2 + cos^{2} (x) + 4 cos (x) = 0

Solve by substitution

- 2 + cos^{2} (x) + 4 cos (x) = 0

Let:

cos (x) = u

- 2 + u^{2} + 4 u = 0

- 2 + u^{2} + 4 u = 0 : u = - 2 + 6, u = - 2 - 6

- 2 + u^{2} + 4 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 4 u - 2 = 0

Solve with the quadratic formula

u^{2} + 4 u - 2 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = - 2

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 2) = 26

4^{2} - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 4^{2} + 8

4^{2} = 16

= 16 + 8

Add the numbers:

16 + 8 = 24

= 24

Prime factorization of

24 : 2^{3} \cdot 3

24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 3

Apply radical rule:

= 2^{2} 2 \cdot 3

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 3

Refine

= 26

u_{1, 2} = \frac{- 4 \pm 2 6}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 4 + 2 6}{2 \cdot 1}, u_{2} = \frac{- 4 - 2 6}{2 \cdot 1}

u = \frac{- 4 + 2 6}{2 \cdot 1} : - 2 + 6

\frac{- 4 + 2 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 + 2 6}{2}

Factor

- 4 + 26 : 2 (- 2 + 6)

- 4 + 26

Rewrite as

= - 2 \cdot 2 + 26

Factor out common term

2

= 2 (- 2 + 6)

= \frac{2 ( - 2 + 6 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 2 + 6

u = \frac{- 4 - 2 6}{2 \cdot 1} : - 2 - 6

\frac{- 4 - 2 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 - 2 6}{2}

Factor

- 4 - 26 : - 2 (2 + 6)

- 4 - 26

Rewrite as

= - 2 \cdot 2 - 26

Factor out common term

2

= - 2 (2 + 6)

= - \frac{2 ( 2 + 6 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (2 + 6)

Negate

- (2 + 6) = - 2 - 6

= - 2 - 6

The solutions to the quadratic equation are:

u = - 2 + 6, u = - 2 - 6

Substitute back

u = cos (x)

cos (x) = - 2 + 6, cos (x) = - 2 - 6

cos (x) = - 2 + 6, cos (x) = - 2 - 6

cos (x) = - 2 + 6 : x = arccos (- 2 + 6) + 2 πn, x = 2 π - arccos (- 2 + 6) + 2 πn

cos (x) = - 2 + 6

Apply trig inverse properties

cos (x) = - 2 + 6

General solutions for

cos (x) = - 2 + 6

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (- 2 + 6) + 2 πn, x = 2 π - arccos (- 2 + 6) + 2 πn

x = arccos (- 2 + 6) + 2 πn, x = 2 π - arccos (- 2 + 6) + 2 πn

cos (x) = - 2 - 6 :

No Solution

cos (x) = - 2 - 6

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (- 2 + 6) + 2 πn, x = 2 π - arccos (- 2 + 6) + 2 πn

Show solutions in decimal form

x = 1.10460 \dots + 2 πn, x = 2 π - 1.10460 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(x)=sin^2(x)+1 ?
The general solution for 4cos(x)=sin^2(x)+1 is x=1.10460…+2pin,x=2pi-1.10460…+2pin