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: 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?

Final answer:

The trigonometric equation is transformed into a quadratic equation by substituting sin θ with x. Then, the quadratic formula is applied to find x, which is then used to find the solutions for θ in radians, considering the periodic nature of the sine function. The solutions are then verified.

Explanation:

To solve the trigonometric equation 2 sin^2 θ − sin θ − 1 = 0, treat it as a quadratic equation by setting x = sin θ. The reformed equation is 2x^2 - x - 1 = 0. Now, factor this equation or use the quadratic formula to find the values of x, and subsequently the values of θ.

Using the quadratic formula:

Therefore, the solutions for x are:

Convert these back into solutions for θ by finding θ such that sin θ = x. Use units of radians for angles and remember to consider the periodic nature of the sine function.

Answer:

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