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Find the solutions in the interval [O,2pi) Sin^2x=2+2cosx I feel kind of dumb any help? 1-cos^2x=2+2cosx 0=cos^2x+2cosx+1
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try this... its explained in details... https://www.symbolab.com/solver/trigonometric-equation-calculator/%5Csin%5E%7B2%7D(x)%3D2%2B2%5Ccos(x)%2C%20%5B0%2C%202%5Cpi%5D/?origin=dym
U subsitution to quadratic formula... nice.
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