Question

How do I solve these trigonometric equations!!?

Answer 1

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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