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 May 5th, 2017, 04:18 AM #1 Newbie   Joined: May 2017 From: Pakistan Posts: 7 Thanks: 0 Answer to these questions ? I've got my exams after 4 days - please solve below questions; thanks. Solve for θ sin 3θ - Sin 2θ - sin θ = 0 Solve for x √3tan x - sec x - 1 = 0 Solve: 1 + cos 2θ = - cosθ Solve sinθ2 - cosθ = 0 Solve 5 sinθ cosθ = 1 Solve 3sin²θ - sinθ = 1/4 Please thanks I will be very thankful. Last edited by skipjack; May 5th, 2017 at 05:29 AM.
May 5th, 2017, 04:32 AM   #2
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 Originally Posted by ahmed786 I've got my exams after 4 days - please solve below questions; thanks. Solve for θ Sin 3θ - Sin 2θ - sin θ = 0
For this one, what values of theta here do you think will cause the statement to be true?

 May 5th, 2017, 05:35 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 0 = sin 3θ - sin 2θ - sin θ = 2cos 2θ sin θ - sin 2θ = 2sin θ (cos 2θ - cos θ) Hence sin θ = 0 (which implies θ = k$\pi$) or 2θ = ±θ + 2k$\pi$, where k is an integer. Can you finish from there? Multiplying √3tan x - 1 = sec x by (1/2)cos x gives cos(x - 2$\pi$/3) = cos($\pi$/3), so x - 2$\pi$/3 = ±$\pi$/3 + 2k$\pi$, where k is an integer. Can you finish from there? 1 + cos 2θ = - cosθ implies 2cos²θ = - cos θ, so cos θ = 0 or -1/2. Can you finish from there? I didn't understand your equation sinθ2 - cosθ = 0. 5 sinθ cosθ = 1 implies sin 2θ = 2/5. What progress can you make from that? 3sin²θ - sinθ = 1/4 implies 3(sin θ + 1/6)(sin θ - 1/2) = 0, so sin θ = - 1/6 or 1/2. Can you finish from there?
 May 5th, 2017, 07:55 AM #4 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 For no1. $\displaystyle \sin3x-\sin2x-\sin x=0$ $\displaystyle -\sin^3x+3\cos^2x \sin x -2 \cos x \sin x- \sin x=0$ $\displaystyle \sin x(4\cos^2x-2 \cos x-\sin^2x-1)=0$ $\displaystyle \sin x=0$ or $\displaystyle 3\cos^2x-2\cos x-\sin^2x -1=0$ $\displaystyle x=2n\pi,\pi+2n\pi$ $\displaystyle 3\cos^2x-2\cos x-1+\cos^2x-1=0$ $\displaystyle 4\cos^2x-2\cos x-2=0$ $\displaystyle (\cos x-1)(4\cos x+2)=0$ $\displaystyle \cos x=1$ or $\displaystyle 4\cos x+2=0$ $\displaystyle x=2n\pi$ or $\displaystyle \cos x=-\frac{1}{2}$ $\displaystyle x=\frac{2\pi}{3}+2n\pi,\frac{4\pi}{3}+2n\pi$ Conclusion, $\displaystyle x=2n\pi,\pi+2n\pi,\frac{2\pi}{3}+2n\pi,\frac{4\pi} {3}+2n\pi$

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