# Solve $\sin(2x)+\cos(x)-2\sin(x)-1=0,$ for $x\in[-\pi,\pi]$

Solve $$\sin(2x)+\cos(x)-2\sin(x)-1=0,\quad x\in[-\pi,\pi].$$

I tried to make this into a quadratic equation so that I could solve for $x$ by converting $\sin(2x)$ into $2\sin(x)\cos(x)$ and then rearranging it somehow. Am I on the right track or not? Both ways, could I be offered a hint to the problem first rather than answers? Thanks very much!

• You're on the right track. All you need to do now is to express cos(x) using sin(x). That you can do using one very important trigonometric formula. – Keen Aug 23 '17 at 12:17
• @Keen Normally I'd agree with you, but I don't think OP even needs to go that far. The expression should just factorise. – T. Linnell Aug 23 '17 at 12:18
• $2 \sin x \cos x+ \cos x - 2\sin x = 2 \sin x(\cos x - 1)+(\cos x -1) = \ldots$ – Kevin Aug 23 '17 at 12:20
• @Kevin thanks! I couldn't think of that somehow... Is it because of lack of experience in solving these problems? – LHC2012 Aug 23 '17 at 12:23
• @InfiniteAccelerator0643 Well, I would put it down to practice makes perfect..the more examples you undertake and the tricks and substitutions you pick up on the way means that you'll smash these types of problems in no time. – Kevin Aug 23 '17 at 13:17

HINT: $$\cos(x)(2\sin(x)+1)-(2\sin(x)+1)=0$$ and this is $$(2\sin(x)+1)(\cos(x)-1)=0$$