Question

Solve the equation on the interval [0,2pi)

Answer 1

The value of θ within the interval is 210 and 330 degrees

Given the equation;

$2sin^2 \theta = 5sin \theta - 3$

• Let P = sinθ

The equation will become;

2P² = 5P - 3

2P² - 5P - 3 = 0

Factorize;

2P² - 6P + P- 3 = 0

2P(P-3) + 1(P -3) =0

2P + 1 = 0 and P - 3  = 0

p = -1/2 and  3

Since P = sinθ

sinθ = -1/2

θ = -30 degres

Since sine is negative in the second 3rd and 4th quadrant

θ = 180 + 30 = 210 degrees

θ2 = 360  - 30 = 330 degrees

Hence the value of θ within the interval is 210 and 330 degrees

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