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

Q: What is the general solution for sin^2(θ)=2cos(θ)+2,0<= θ<= 2pi ?

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

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sin^2(θ)=2cos(θ)+2,0<= θ<= 2pi

Solution

sin^{2} (θ) = 2 cos (θ) + 2, 0 \leq θ \leq 2 π

Solution

θ = π

Degrees

θ = 18 0^{\circ}

Solution steps

sin^{2} (θ) = 2 cos (θ) + 2, 0 \leq θ \leq 2 π

Subtract

2 cos (θ) + 2

from both sides

sin^{2} (θ) - 2 cos (θ) - 2 = 0

Rewrite using trig identities

- 2 + sin^{2} (θ) - 2 cos (θ)

Use the Pythagorean identity:

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

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

= - 2 + 1 - cos^{2} (θ) - 2 cos (θ)

Simplify

= - cos^{2} (θ) - 2 cos (θ) - 1

- 1 - cos^{2} (θ) - 2 cos (θ) = 0

Solve by substitution

- 1 - cos^{2} (θ) - 2 cos (θ) = 0

Let:

cos (θ) = 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 (θ)

cos (θ) = - 1

cos (θ) = - 1

cos (θ) = - 1, 0 \leq θ \leq 2 π : θ = π

cos (θ) = - 1, 0 \leq θ \leq 2 π

General solutions for

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = π

Combine all the solutions

θ = π

Popular Examples

sin(x)=-0.3

sin (x) = - 0.3

-sin(x)+cos(x)=0,-pi<= x<= pi

- sin (x) + cos (x) = 0, - π \leq x \leq π

2sqrt(3)cos(x)-4sin(x)cos(x)=0

23 cos (x) - 4 sin (x) cos (x) = 0

sinh(x)=2tanh(x)

sinh (x) = 2 tanh (x)

sin^2(x)=2-2cos(x),0<= x<= 2pi

sin^{2} (x) = 2 - 2 cos (x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

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