Answer:
Solution are
i) \(x = 0, \pi, \frac{3\pi}{2} , 2\pi\)
Step-by-step explanation:
1) (sinx)(cosx) =0
sin2x/2 =0
2x =0, 2pi, 3pi, 4pi
Hence \(x = 0, \pi, \frac{3\pi}{2} , 2\pi\)
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2) \(cos2x + 2 cos x + 1 = 0\\2cos^2 x +2cosx +1 =0\\(cosx+1)^2 =0\\cosx =-1\\x=\pi\)
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