+9479 4 sin2x = 4 cos x + 5
By the Pythagorean identity, sin2x = 1 - cos2x so....
4( 1 - cos2x ) = 4 cos x + 5
Distribute the 4 to the terms in parenthesees.
4 - 4 cos2x = 4 cos x + 5
Add 4 cos2 x to both sides of the equation.
4 = 4 cos2x + 4 cos x + 5
Subtract 4 from both sides of the equation.
0 = 4 cos2x + 4 cos x + 1
Split 4 cos x into two terms that we can use to factor by grouping.
0 = 4 cos2x + 2 cos x + 2 cos x + 1
0 = 2 cos x ( 2 cos x + 1 ) + 1( 2 cos x + 1 )
0 = ( 2 cos x + 1 )( 2 cos x + 1 )
0 = ( 2 cos x + 1 )2
Take the square root of both sides.
0 = 2 cos x + 1
Subtract 1 from both sides.
-1 = 2 cos x
Divide both sides by 2 .
-1/2 = cos x
The angle that has a cosine of -1/2 in the interval [0, pi) is 2 pi / 3
x = 2 pi / 3
