solve 2sin2x−sinx=0 in the interval 0°<x<90° ?
- Jeff AaronLv 71 month ago
nyphdinmd is correct, but note that dividing by sin(x) only works if sin(x) is not zero. However, since 0°<x<90°, then sin(x) can't be zero, so nyphdinmd is correct that the only solution is cos(x) = 1/4, so x =~ 75.522 degrees.
- nyphdinmdLv 71 month ago
write sin(2x) = sin(x +x) = 2sin(x)cos(x) then
2*sin(2x) -sin(x) = 4sin(x)cos(x) - sin(x) = 0 divide by sin(x)
4cos(x) - 1 = 0 --> cos(x) = 1/4 --> x = acos(1/4) = 75.522 deg