What is the general solution for -4cos^{(2)}(x)+4sqrt(2)sin(x)+6=0 ?

The general solution for -4cos^{(2)}(x)+4sqrt(2)sin(x)+6=0 is x=(5pi)/4+2pin,x=(7pi)/4+2pin

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Popular Trigonometry >

-4cos^{(2)}(x)+4sqrt(2)sin(x)+6=0

Solution

- 4 cos^{(2)} (x) + 42 sin (x) + 6 = 0

Solution

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos^{(2)} (x) + 42 sin (x) + 6 = 0

Rewrite using trig identities

6 - 4 cos^{2} (x) + 4 sin (x) 2

Use the Pythagorean identity:

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

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

= 6 - 4 (1 - sin^{2} (x)) + 4 sin (x) 2

Simplify

6 - 4 (1 - sin^{2} (x)) + 4 sin (x) 2 : 4 sin^{2} (x) + 42 sin (x) + 2

6 - 4 (1 - sin^{2} (x)) + 4 sin (x) 2

= 6 - 4 (1 - sin^{2} (x)) + 42 sin (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

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

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

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

= 6 - 4 + 4 sin^{2} (x) + 4 sin (x) 2

Subtract the numbers:

6 - 4 = 2

= 4 sin^{2} (x) + 42 sin (x) + 2

= 4 sin^{2} (x) + 42 sin (x) + 2

2 + 4 sin^{2} (x) + 4 sin (x) 2 = 0

Solve by substitution

2 + 4 sin^{2} (x) + 4 sin (x) 2 = 0

Let:

sin (x) = u

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

2 + 4 u^{2} + 4 u 2 = 0 : u = - \frac{2}{2}

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

Write in the standard form

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

4 u^{2} + 42 u + 2 = 0

Solve with the quadratic formula

4 u^{2} + 42 u + 2 = 0

Quadratic Equation Formula:

For

a = 4, b = 42, c = 2

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

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

(42)^{2} - 4 \cdot 4 \cdot 2 = 0

(42)^{2} - 4 \cdot 4 \cdot 2

(42)^{2} = 4^{2} \cdot 2

(42)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (2)^{2}

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 4^{2} \cdot 2

4 \cdot 4 \cdot 2 = 32

4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 32

= 4^{2} \cdot 2 - 32

4^{2} \cdot 2 = 32

4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

Multiply the numbers:

16 \cdot 2 = 32

= 32

= 32 - 32

Subtract the numbers:

32 - 32 = 0

= 0

u_{1, 2} = \frac{- 4 2 \pm 0}{2 \cdot 4}

u = \frac{- 4 2}{2 \cdot 4}

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

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 2}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4 2}{8}

Cancel the common factor:

4

= - \frac{2}{2}

u = - \frac{2}{2}

The solution to the quadratic equation is:

u = - \frac{2}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{2}{2}

sin (x) = - \frac{2}{2}

sin (x) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{2}{2}

General solutions for

sin (x) = - \frac{2}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -4cos^{(2)}(x)+4sqrt(2)sin(x)+6=0 ?
The general solution for -4cos^{(2)}(x)+4sqrt(2)sin(x)+6=0 is x=(5pi)/4+2pin,x=(7pi)/4+2pin