What is the general solution for sin^2(x)=2+2cos(x),(0,2pi) ?

The general solution for sin^2(x)=2+2cos(x),(0,2pi) is x=pi

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

Solution

sin^{2} (x) = 2 + 2 cos (x), (0, 2 π)

Solution

x = π

Degrees

x = 18 0^{\circ}

Solution steps

sin^{2} (x) = 2 + 2 cos (x), (0, 2 π)

Subtract

2 + 2 cos (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

cos (x) = u

- 1 - u^{2} - 2 u = 0

- 1 - u^{2} - 2 u = 0 : u = - 1

- 1 - u^{2} - 2 u = 0

Write in the standard form

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

- u^{2} - 2 u - 1 = 0

Solve with the quadratic formula

- u^{2} - 2 u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 (- 1) (- 1) = 0

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 ( - 1 )}

u = \frac{- ( - 2 )}{2 ( - 1 )}

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

\frac{- ( - 2 )}{2 ( - 1 )}

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1, (0, 2 π) : x = π

cos (x) = - 1, (0, 2 π)

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

Solutions for the range

(0, 2 π)

x = π

Combine all the solutions

x = π

Popular Examples

1-sin(θ)2=(1-sin(θ))(1-sin(θ))

1 - sin (θ) 2 = (1 - sin (θ)) (1 - sin (θ))

cos(x)= 22/24

cos (x) = \frac{22}{24}

sin(x)=(1.2)/(1.6)

sin (x) = \frac{1.2}{1.6}

2cos^2(θ)+3cos(θ)=-1

2 cos^{2} (θ) + 3 cos (θ) = - 1

sin(x)=(14.9)/(19.9725)

sin (x) = \frac{14.9}{19.9725}

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)=2+2cos(x),(0,2pi) ?
The general solution for sin^2(x)=2+2cos(x),(0,2pi) is x=pi