Consider the given equation,
2cos2θ−√3sinθ+1=0
2(1−sin2θ)−√3sinθ+1=0 [∵cos2θ=1−sin2θ]
2−2sin2θ−√3sinθ+1=0
2sin2θ+√3sinθ−3=0
sinθ=−√3±√3−4×2×−32×2 (use quadratic formula)
sinθ=−√3±√3+242×2
sinθ=−√3±3√34
∴sinθ=√32,−√3
Solve the equation, 2cos2θ+3sinθ=0
Or
Solve the equation, sec22x=1−tan2x