Squaring on both sides $$\sin^2 x +\cos^2 x + 2\sin x \cos x=1+\sin^2 x \cos^2 x +2\sin x \cos x$$ $$\sin^2 x\cos^2 x=0$$ $$\sin 2x=0$$
I feel this answer is wrong because the answer is $n\pi + (-1)^n\pi /2$
What am I doing wrong?
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Sign up to join this communitySquaring on both sides $$\sin^2 x +\cos^2 x + 2\sin x \cos x=1+\sin^2 x \cos^2 x +2\sin x \cos x$$ $$\sin^2 x\cos^2 x=0$$ $$\sin 2x=0$$
I feel this answer is wrong because the answer is $n\pi + (-1)^n\pi /2$
What am I doing wrong?
Avoid squaring as it immediately introduces When do we get extraneous roots?
Your solution includes $$\sin x+\cos x=-(1+\sin x\cos x)$$ as well
Use $$a+b=1+ab\iff(1-a)(1-b)=0$$
To avoid When do we get extraneous roots?
set $$\sin x+\cos x=u,u^2=?$$
$$2u=2+u^2-1\implies(u-1)^2=?$$