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The equation goes as follows: $$\sin x +\cos x = 1 + \sin x \cos x$$

and here is how I solved it:

$$(\sin x+\cos x)^2=(1+\sin x\cos x)^2$$ $$\sin^2x+2\sin x\cos x+\cos^2x=1+2\sin x\cos x+2\sin^2x\cos^2x$$ $$2\sin^2x\cos^2x=0$$

$$ \cos x=0; \sin x=0$$ $$x_1=\pi/2 +k\pi$$ $$x_2=k\pi$$

Where did it go wrong? Thanks in advance!

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    $\begingroup$ Why did you get a $2\sin^2(x)\cos^2(x)$? Shouldn't the coefficient be $1$? $\endgroup$ – Michael Burr Jul 1 '15 at 21:23
  • $\begingroup$ Why do you believe that you made a mistake? Make sure to plug in your answers into the original expression because squaring may have introduced extra solutions. $\endgroup$ – Michael Burr Jul 1 '15 at 21:23
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Hint.

Your work is correct, but incomplete. You have to verifiy the solutions because squaring you can introduce improper solutions and in your case these are the solutions for $k$ odd.

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The easiest way to solve this is just to factor everything like this:

$$\sin(x)+\cos(x)-\sin(x)\cos(x)-1=(\sin(x)-1)(\cos(x)-1)=0$$ So you can see immediatly that you have solutions when $\cos(x)=1$ or $\sin(x)=1$, thus $x=2k\pi$ or $x=\frac{\pi}{2}+2k\pi$

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    $\begingroup$ One quibble: $\sin x + \cos x - \sin x\cos x - 1 = (\sin x - 1)(1 - \cos x)$. $\endgroup$ – N. F. Taussig Jul 2 '15 at 9:00
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On your second line, the third term on the right hand side shouldn't have a 2 in front of it.

When you square a binomial, you get the general form $$(a+b)^2 = a^2 + 2ab + b^2$$

So in your case $$(1 + \sin(x)\cos(x))^2 = 1 + 2\sin(x)\cos(x) + \sin^2(x)\cos^2(x)$$

From there, you should be able to solve for x quite easily

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    $\begingroup$ This is a mistake in the OP's solution, but observe that it doesn't change the answer that the OP developed. $\endgroup$ – Michael Burr Jul 1 '15 at 21:25
  • $\begingroup$ I wish my mistakes had a similar null effect. $\endgroup$ – marty cohen Jul 1 '15 at 22:18
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You have a mistake in calculating $(A+B)^2=A^2+2AB+B^2$ (though this mistake does not affect the result.)

In addition to this, $\cos x=0\ \text{or}\ \sin x=0$ is just a necessary condition. Because you squared the both sides, you need to check if they are sufficient. (For example, $x=3\pi/2$ is not a solution.)

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  • $\begingroup$ You are right that a mistake was made. However, the mistake did not affect the result since its effect was coming up with $2f(x) = 0$ instead of the correct $f(x) = 0$. Good catch though. $\endgroup$ – marty cohen Jul 1 '15 at 22:17
  • $\begingroup$ @martycohen: You are right. When I first wrote my answer, I wrote only the second paragraph. Then, I began to think that adding the first paragraph may be better :) (maybe I should have mentioned that that mistake does not affect the result. so edited) $\endgroup$ – mathlove Jul 1 '15 at 22:23

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