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Sergio [31]
3 years ago
5

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac

es where appropriate. If there is no solution, enter NO SOLUTION.)
2 sin^2 θ − sin θ − 1 = 0 ...?
Mathematics
2 answers:
worty [1.4K]3 years ago
8 0
So here is the answer to the given equation above.
To find the answer, we can just factor this one. 
<span>(2sinθ+1)(sinθ−1)=0</span><span>sinθ = <span><span>−1/</span>2</span>→θ = <span>{<span><span><span>−π/</span>6 </span>+ 2kπ, <span><span>7π/</span>6 </span>+ 2kπ</span>}</span></span><span><span>sinθ=1→θ=<span>{<span><span>π/2 </span>+ 2kπ</span>}
Hope this answers your question. Let me know if you need more help next time. Have a great day!</span></span></span>
Gnoma [55]3 years ago
5 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?

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