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The following equation involves trigonometric equations quadratic in form. Solve the equation on the interval [0,2π).

The following equation involves trigonometric equations quadratic in form. Solve the-example-1

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3 sin ^2 x = 8 sin x -5

Move the terms to the left hand side

3 sin ^2 x - 8 sin x -5 =0

Let m = sin x

3 m^2 -8m -5 =0

Factor

(3m-5)(m-1) =0

Using the zero product property

3m-5 =0 m-1 =0

3m =5 m=1

m = 5/3 m=1

Now replace m with sin x

sin x = 5/3 sin x =1

Taking the inverse sin of each side

sin ^-1 ( sin x) = sin ^-1( 5/3) sin ^-1(sin x) = sin ^-1 (1)

x = no solution x = pi/2 + 2pi n where n is an integer

We limit our solution to the interval [0,2pi)

x = pi/2

User James Tauber
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