Solving a Trigonometric Equation?

Question

sin^2 = 6(cos(-θ) + 1) ; interval, 0≤θ<2π

Answer 1

Given equation : sin² θ= 6(cos(-θ) + 1)

sin²θ= 6(cos(-θ) + 1)

Substitute : cos(-θ) = cosθ

sin² θ = 6(cosθ + 1)

Substitute : sin² θ = 1 - cos²θ

1 - cos²θ = 6(cosθ + 1)

cos²θ+ 6cosθ + 6 - 1 = 0

cos²θ+ 6cosθ + 5 = 0

cos²θ+ cosθ + 5cosθ + 5 = 0

cosθ ( cosθ + 1 ) + 5 ( cosθ + 1 ) = 0

( cosθ + 1 )( cosθ + 5 ) = 0

By using zero product property : If AB = 0 then A = 0 , B = 0

( cosθ + 1 ) = 0  and  ( cosθ + 5 ) = 0

cosθ = -1  and  cosθ = -5

The condition for cosθ is -1 ≤ cosθ  ≤1.

So cosθ = -5 is invalid.

cosθ = -1 is valid

cosθ = -1

cosθ = cos π

General solution for cosine is : 2nπ ± θ , where n = 0,1,2,3 .....

θ = 2nπ ± π

If n = 0 then θ = 2(0)π ± π = π

If n = 1 then θ = 2(1)π ± π = 3π

3π is out of given range 0≤θ<2π.

So n = 1 and above are not solutions for given range.

So

θ = π = 180°

The solution for givenequation is θ = 180°