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4sin(x)^2 + 1=8cos(x) Solve for x
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\[4(\sin x)^{2}+1=8cosx\]
convert sin^2x to cos^2x first
4(sin^2x) +1 =8cosx 4(1-cos^2x) +1 =8cosx 4cos^2x +8cosx-5=0
thanks
4cos^2x +10cosx-2cosx-5=0 2cosx(2cosx+5)-1(2cosx+5)=0 (2cosx-1)(2cosx+5)=0
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cosx <=1 hence cosx=1/2 x= 2n(pi)+- (pi/3) where n is any integer
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