\({2cos^2x-sinx-1} = 0\)
The answers are π/6, 5π/6, and -π/2, how should I get these answers?
To solve 2cos2(x) - sin(x) - 1 = 0 use the identity: cos2(x) = 1 - sin2(x):
---> 2[ 1 - sin2(x) ] - sin(x) - 1 = 0
2 - 2sin2(x) - sin(x) - 1 = 0
-2sin2(x) - sin(x) + 1 = 0
2sin2(x) + sin(x) - 1 = 0
[ 2sin(x) - 1 ][ sin(x) + 1 ] = 0
So: either 2sin(x) - 1 = 0 ---> 2sin(x) = 1 ---> sin(x) = ½ ---> x = sin-1(½)
or sin(x) + 1 = 0 ---> sin(x) = -1 ---> x = sin-1(-1)
From here, you can get the answers ...