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solve: $\sin x=-\cos x$ in $ [-\pi,\pi]$ Can someone mainly just explain the interval part as I know how to solve the other part.

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    What do you mean by asking for the ‘interval part’ of the solution? – 2017-01-14
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    Well I know the Radians, but I just need some help with the domain part – 2017-01-14
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    If you are not clear enough, I can't possibly guess what troubles you... What is the ‘domain part’? The exercise asks to solve the trig equation if you know that $x$ is a number between (or equal to) $- \pi$ and $\pi$. – 2017-01-14
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    Well I can get the solution for TanX=-1, I just need help putting them in the domain [-pi,pi] – 2017-01-14

3 Answers 3

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Hint:

$$\sin x=-\cos x\implies \tan x=-1$$

Fill in details: why can we divide by $\;\cos x\;$ ? At what points does the tangent function equal $\;-1\;$ ?

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The equation can be written as $\cos(x)=\cos(x+\frac{\pi}{2})$ which gives

$$x=-x-\frac{\pi}{2}+2k\pi.$$ or $$x=-\frac{\pi}{4}+k\pi$$

as $x\in[-\pi,\pi]$, the solutions are

$$\{\frac{-\pi}{4},\frac{3\pi}{4}\}$$ with $k=0,1$.

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In order to solve the equality $$\sin (x) + \cos (x) = 0,$$ $x \in [-\pi,\pi]$, one can square both sides of the equation and use the identity $$\sin ^2 (x) + \cos ^2(x) = 1.$$