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$\sin x+1=\cos x, x\in[-\pi,\pi]$

Hi I'm confused about a trig question. The question asks to solve using a half angle identity and in the solution it states to subtract 2π from 3π/2 resulting in one of the solutions being -π/2 along with another solution of zero. After using the half angle identity I got -1=tanx/2 and from there in Q2 I found x to be 3π/2 and in Q4 I found x to be 7π/2. Neither of these are in the domain so the solution says to subtract 2π from 3π/2. Would the period be 2π since it is a half angle identity, I'm confused how you know the period is 2π since it is tan?
Thanks

Martin Sleziak
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melanie
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1 Answers1

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$(\sin x+1=\cos x)$ iff $(1-\cos x=-\sin x)$ iff $(2\sin^2(x/2)=\sin x=-2\sin (x/2) \cos (x/2))$ iff $(\sin (x/2)=0$ or $0\ne \sin (x/2)=-\cos (x/2)).$ Now $x\in [-\pi,\pi]$ iff $x/2\in [-\pi/2,\pi/2]$ so (1)...$\sin (x/2)=0$ iff $x/2=0$ iff $x=0$, while (2)... $(0\ne \sin (x/2)=-\cos (x/2))$ iff $ x/2=\pi/4$ iff $x=\pi/2.$

DanielWainfleet
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