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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, $\displaystyle x=2n\pi,\pi+2n\pi,\frac{2\pi}{3}+2n\pi,\frac{4\pi} {3}+2n\pi$ Tags answer, questions, trignometry Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post wannamonroe Pre-Calculus 5 April 15th, 2016 10:44 AM mohan Probability and Statistics 3 December 22nd, 2014 12:07 PM Shpeck Advanced Statistics 2 June 30th, 2012 07:18 PM ewout Probability and Statistics 2 November 4th, 2010 06:05 PM alti Algebra 1 September 21st, 2009 05:18 AM

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