Solve -2cos^2x-3sinx=0 for 0<x<2pi?

Question

Trig. homework. Thank you!

Answer 1

The trigonometric equation is .

                                   

.

Therefore, solution of the equation is 

.

Answer 2

The trigonometric equation is - 2cos²(x) - 3sin(x) = 0, 0 < x < 2π and solve for x.

- 2[ 1 - sin²(x) ] - 3sin(x) = 0

- 2 + 2sin²(x) - 3sin(x) = 0

2sin²(x) - 3sin(x) - 2 = 0

2sin²(x) - 4sin(x) + sin(x) - 2 = 0

2sin(x) [ sin(x) - 2 ] + 1 [ sin(x) - 2 ] = 0

[ 2sin(x) + 1 ] [ sin(x) - 2 ] = 0

2sin(x) = - 1 or sin(x) = 2.

sin(x) = - 1/2 or sin(x) = 2.

x = sin^{- 1}(- 1/2) or x = sin^{- 1}(2).

The function sin(x) has a period of 2π, first find all solutions in the interval (0, 2π).

The function sin(x) is negative in third and fourth quadrant.

In third Quadrant, π ≤ x ≤ 3π/2.

- 1/2 = - sin (π/6) = sin (π + π/6) = sin (7π/6).

In fourth Quadrant, 3π/2 ≤ x ≤ 2π.

- 1/2 = - sin (π/6) = sin (2π - π/6) = sin (11π/6).

So the solutions are

x = sin^{- 1}(- 1/2) -----> x = sin^-1[sin (7π/6)] -------> x = 7π/6.

x = sin^{- 1}(- 1/2) -----> x = sin^-1[sin (11π/6)] -------> x = 11π/6 and

x = sin^{- 1}(2).

The solutions x = 7π/6, x = 11π/6 and x = sin^{- 1}(2) in the interval (0, 2π).