Answered

Solve the following equations for all solutions of x


2sin^2x+3cos-3=0

Verified Answer :

syed514

Remember that sin^2x + cos^2x = 1 so sin^2x = 1 - cos^2x
2sin^2 x + 3cosx - 3 = 0
2(1 - cos^2x) + 3cosx - 3 = 0
2cos^2x - 3cosx + 1 = 0
This is just a quadratic equation in cosx
So let u = cosx
2u^2 - 3u + 1 = 0
This factors as
(2u-1)(u-1) = 0
So 2u = 1 and u = 1 or 2cosx = 1 and cosx = 1
cosx = 1 when x = 0cosx = 1/2 when x = pi/3 and 5pi/3
So x = {0, pi/3, 5pi/3} is the solution set for your equation.

HomertheGenius

2 sin² x + 3 cos x - 3 = 0
2 ( 1 - cos² x ) + 3 x - 3 = 0
2 - 2 cos² x + 3 cos x - 3 = 0
- 2 cos² x + 3 cos x - 1 = 0   / * ( - 1 )
2 cos² x - 3 cos x + 1 = 0
Substitution:  u = cos x
2 u² - 3 u + 1 = 0
2 u² - 2 u - u + 1 = 0
2 u ( u - 1 ) - ( u - 1 ) = 0
( u - 1 ) ( 2 u - 1 ) = 0,    u 1 = 1,  u 2 = 1/2
cos x = 1,    cos x = 1/2
Answer:
x 1 = k π,   x 2 = π / 3 + 2 k π,   x 3 = 5 π / 3 + 2 k π,  k ∈ Z

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