Solve for x : 4sin^2x-4cosx=1
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4(sinx)²-4cosx=1
- (sinx)²+(cosx)²=1 ⇒(sinx)²=1-(cosx)²
⇒4(1-(cosx)²)-4cosx=1
⇒4-4(cosx)²-4cosx=1
⇒4(cosx)²+4cosx-3=0
⇒4(cosx)²-2cosx+6cosx-3=0
⇒2cosx(2cosx-1)+3(2cosx-1)=0
⇒(2cosx-1)(2cosx+3)=0
⇒cosx=1/2
∴x=π/6+nπ.
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