+318 Solve each trigonmetric equation in the interval [0, 2pi) by first squarring both sides.
sq rt(3) cos x=sin x+1
solution set??????
type an exact answer, using pi as needed.
+23252 Solve by squaring both sides:
sqrt(3)·cos(x) = sin(x) + 1
sqrt(3)·cos(x) ]2 = [ sin(x) + 1 ]2
3·cos2(x) = sin2(x) + 2·sin(x) + 1
Since cos2(x) = 1 - sin2(x)
3·( 1 - sin2(x) ) = sin2(x) + 2·sin(x) + 1
3 - 3·sin2(x) = sin2(x) + 2·sin(x) + 1
0 = 4sin2(x) + 2sin(x) - 2
0 = 2sin2(x) + sin(x) - 1
0 = ( 2sin(x) - 1 )( sin(x) + 1 )
So: either 2sin(x) - 1 = 0 or sin(x) + 1 = 0
2sin(x) = 1 sin(x) = -1
sin(x) = ½ x = sin-1(1)
x = sin-1( ½ )
Now, you'll need to finish this ...
