+318 Solve each trigonmetric equation in the interval (0, 2pi) by first squarring both sides.
(SQ RT 3) sin x=cos x+1
PLS HELP W THE ANSWER
SOLUTION SET=????
+129899 √3sin x = cos x+ 1 square both sides
3sin^2 x = cos^2x + 2cos x + 1
3 ( 1 - cos^2x) = cos^2x + 2cosx + 1
3 - 3cos^2 x = cos^2x + 2cos x + 1 rearrange as
4cos^2 x + 2cos x - 2 = 0 divide through by 2
2cos^2 x + cos x - 1 = 0 factor as
(2cos x - 1) ( cos x + 1) = 0
So either
2cos x - 1 = 0
2cos x = 1
cos x = 1/2
And this is true at x = 60° and x = 360 - 60 = 300°
or
cos x + 1 = 0
cos x = - 1
And this is true at x = 180°
Because we squred both sides we need to check these solutions
√3 sin (60) = 1 + cos (60)
√3 (√3/2) = 1 + 1/2
3/2 = 3/2 true
√3 sin (180) = 1 + cos (180)
0 = 1 - 1
0 = 0 true
√3 sin (300) = 1 + cos (300)
√3 (-1/2) = 1 + 1/2
-√3/2 = 3/2 not true
So...the only solutions are x = 60° and x = 180°
