Question

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
2 sin^2 θ − sin θ − 1 = 0 ...?

Answer 1

So here is the answer to the given equation above.
To find the answer, we can just factor this one. 
(2sinθ+1)(sinθ−1)=0sinθ = −1/2→θ = {−π/6 + 2kπ, 7π/6 + 2kπ}sinθ=1→θ={π/2 + 2kπ}
Hope this answers your question. Let me know if you need more help next time. Have a great day!

Answer 2

Answer: 30°, 300° and 330°

Step-by-step explanation:

This is a quadratic equation in trigonometry format.

Given 2 sin^2 θ − sin θ − 1 = 0

Let a constant 'k' = sin θ...(1)

The equation becomes

2k²-k-1 =0

Factorizing the equation completely we have,

(2k²-2k)+(k-1) = 0

2k(k-1)+1(k-1)=0

(2k+1)(k-1)=0

2k+1=0 and k-1=0

2k = -1 and k=1

k=-1/2 and 1

Substituting the value of k into equation 1 to get θ

sin θ = 1

θ = arcsin1

θ = 90°

Similarly

sin θ = -1/2

θ = arcsin-1/2

θ = -30°

This angle is negative and falls in the 3rd and 4th quadrant

In the third quadrant, θ = 270 +30 = 300° and

in the 4th quadrant, θ = 360 - 30° = 330°

Therefore the values of θ are 30°, 300° and 330°

I hope you find this helpful?