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

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?

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step-by-step explanation:

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