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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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 $$a+b=1+ab\iff(1-a)(1-b)=0$$
answered Nov 20 '19 at 14:10
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$\begingroup$ If that’s the case, then $\cos x$ and $\sin x$ will simultaneously be 1, which isn’t possible $\endgroup$ – Aditya Nov 20 '19 at 14:46
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$\begingroup$ @Aditya, if $$p\cdot q=0,$$ what can we conclude? $\endgroup$ – lab bhattacharjee Nov 20 '19 at 14:48
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$\begingroup$ Assuming a and b are $sin x $ and $cos x$, $ab=0$ should be that $x=\pi , 0, \pi/2$ $\endgroup$ – Aditya Nov 20 '19 at 14:54
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1$\begingroup$ @Aditya, We need $$1-\sin x=0$$ or $$1-\cos x=0$$ right? Why are you saying simultaneously $$1$$ $\endgroup$ – lab bhattacharjee Nov 20 '19 at 14:55
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1$\begingroup$ @Aditya, $$\cos x=1,x=2m\pi$$ right? Check if that satisfies the given equation $\endgroup$ – lab bhattacharjee Nov 20 '19 at 15:05
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To avoid When do we get extraneous roots?
set $$\sin x+\cos x=u,u^2=?$$
$$2u=2+u^2-1\implies(u-1)^2=?$$
answered Nov 20 '19 at 14:52
