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# Solve the following equations: (i) $\mathrm{cos}\mathrm{\theta }+\mathrm{cos}2\mathrm{\theta }+\mathrm{cos}3\mathrm{\theta }=0$ (ii) $\mathrm{cos}\mathrm{\theta }+\mathrm{cos}3\mathrm{\theta }-\mathrm{cos}2\mathrm{\theta }=0$ (iii) $\mathrm{sin}\mathrm{\theta }+\mathrm{sin}5\mathrm{\theta }=\mathrm{sin}3\mathrm{\theta }$ (iv) $\mathrm{cos}\mathrm{\theta }\mathrm{cos}2\mathrm{\theta }\mathrm{cos}3\mathrm{\theta }=\frac{1}{4}$ (v) $\mathrm{cos}\mathrm{\theta }+\mathrm{sin}\mathrm{\theta }=\mathrm{cos}2\mathrm{\theta }+\mathrm{sin}2\mathrm{\theta }$ (vi) $\mathrm{sin}\mathrm{\theta }+\mathrm{sin}2\mathrm{\theta }+\mathrm{sin}3=0$ (vii) $\mathrm{sin}\mathrm{\theta }+\mathrm{sin}2\mathrm{\theta }+\mathrm{sin}3\mathrm{\theta }+\mathrm{sin}4\mathrm{\theta }=0$ (viii) $\mathrm{sin}3\mathrm{\theta }-\mathrm{sin}\mathrm{\theta }=4{\mathrm{cos}}^{2}\mathrm{\theta }-2$ (ix) $\mathrm{sin}2\mathrm{\theta }-\mathrm{sin}4\mathrm{\theta }+\mathrm{sin}6\mathrm{\theta }=0$

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Solution

## (i) $\mathrm{cos}\theta +\mathrm{cos}2\theta +\mathrm{cos}3\theta =0$ Now, $\left(\mathrm{cos}\theta +\mathrm{cos}3\theta \right)+\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}\left(\frac{4\theta }{2}\right)\mathrm{cos}\left(\frac{2\theta }{2}\right)+\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}2\theta \mathrm{cos}\theta +\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2\theta \left(2\mathrm{cos}\theta +1\right)=0$ $⇒\mathrm{cos}2\theta =0$ or, $2\mathrm{cos}\theta +1=0$ $⇒\mathrm{cos}2\theta =\mathrm{cos}\frac{\mathrm{\pi }}{2}$ or $\mathrm{cos}\theta =-\frac{1}{2}=\mathrm{cos}\frac{2\mathrm{\pi }}{3}$ $⇒2\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2}$, $n\in Z$ or $\theta =2m\pi ±\frac{2\mathrm{\pi }}{3},m\in Z\phantom{\rule{0ex}{0ex}}$ $⇒\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{4},n\in Z$ or $\theta =2m\pi ±\frac{2\mathrm{\pi }}{3},m\in Z$ (ii) $\left(\mathrm{cos}\theta +\mathrm{cos}3\theta \right)-\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}\left(\frac{4\theta }{2}\right)\mathrm{cos}\left(\frac{2\theta }{2}\right)-\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}2\theta \mathrm{cos}\theta -\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2\theta \left(2\mathrm{cos}\theta -1\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{cos}2\theta =0$ or $2\mathrm{cos}\theta -1=0$ $⇒\mathrm{cos}2\theta =\mathrm{cos}\frac{\mathrm{\pi }}{2}$ or $\mathrm{cos}\theta =\frac{1}{2}⇒\mathrm{cos}\theta =\mathrm{cos}\frac{\mathrm{\pi }}{3}$ $⇒2\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in Z$ or $\theta =2m\pi ±\frac{\mathrm{\pi }}{3},m\in Z$ $⇒\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{4},n\in Z$ or $\theta =2m\pi ±\frac{\mathrm{\pi }}{3},m\in Z$ (iii) $\mathrm{sin}\theta +\mathrm{sin}5\theta =\mathrm{sin}3\theta$ $⇒2\mathrm{sin}\left(\frac{6\theta }{2}\right)\mathrm{cos}\left(\frac{4\theta }{2}\right)=\mathrm{sin}3\theta \phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}3\theta \mathrm{cos}2\theta =\mathrm{sin}3\theta \phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}3\theta \mathrm{cos}2\theta -\mathrm{sin}3\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}3\theta \left(2\mathrm{cos}2\theta -1\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{sin}3\theta =0$ or $\left(2\mathrm{cos}2\theta -1\right)=0$ $⇒\mathrm{sin}3\theta =\mathrm{sin}0$ or $\mathrm{cos}2\theta =\frac{1}{2}=\mathrm{cos}\frac{\mathrm{\pi }}{3}$ $⇒3\theta =n\pi$ or $2\theta =2m\pi ±\frac{\mathrm{\pi }}{3}$ $⇒$ $\theta =\frac{n\mathrm{\pi }}{3},n\in Z$ or $\theta =m\mathrm{\pi }±\frac{\mathrm{\pi }}{6},m\in \mathit{}Z$ $\left(\mathrm{iv}\right)\mathrm{cos}\theta \mathrm{cos}2\theta \mathrm{cos}3\theta =\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\left[\frac{\mathrm{cos}\left(\theta +2\theta \right)+\mathrm{cos}\left(2\theta -\theta \right)}{2}\right]\mathrm{cos}3\theta =\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒2\left[\mathrm{cos}3\theta +\mathrm{cos}\theta \right]\mathrm{cos}3\theta =1\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{cos}}^{2}3\theta +2\mathrm{cos}\theta \mathrm{cos}3\theta -1=0\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{cos}}^{2}3\theta -1+2\mathrm{cos}\theta \mathrm{cos}3\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}6\theta +\mathrm{cos}4\theta +\mathrm{cos}2\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}6\theta +\mathrm{cos}2\theta +\mathrm{cos}4\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}4\theta \mathrm{cos}2\theta +\mathrm{cos}4\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}4\theta \left(2\mathrm{cos}2\theta +1\right)=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}4\theta =0\mathrm{or}2\mathrm{cos}2\theta +1=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}4\theta =0\mathrm{or}\mathrm{cos}2\theta =\frac{-1}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}4\theta =\mathrm{cos}\frac{\mathrm{\pi }}{2}\mathrm{or}\mathrm{cos}2\mathrm{\theta }=\mathrm{cos}\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒4\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in Z\mathrm{or}2\theta =2m\mathrm{\pi }±\frac{2\mathrm{\pi }}{3},m\in Z\phantom{\rule{0ex}{0ex}}⇒\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{8},n\in Z\mathrm{or}\theta =m\mathrm{\pi }±\frac{\mathrm{\pi }}{3},m\in Z\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (v) $\mathrm{cos}\theta +\mathrm{sin}\theta =\mathrm{cos}2\theta +\mathrm{sin}2\theta$ $⇒\mathrm{cos}\theta -\mathrm{cos}2\theta =\mathrm{sin}2\theta -\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}⇒-2\mathrm{sin}\left(\frac{3\theta }{2}\right)\mathrm{sin}\left(\frac{-\theta }{2}\right)=2\mathrm{sin}\left(\frac{\theta }{2}\right)\mathrm{cos}\left(\frac{3\theta }{2}\right)\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\left(\frac{3\theta }{2}\right)\mathrm{sin}\left(\frac{\theta }{2}\right)=2\mathrm{sin}\left(\frac{\theta }{2}\right)\mathrm{cos}\left(\frac{3\theta }{2}\right)\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\left(\frac{\theta }{2}\right)\left[\mathrm{sin}\left(\frac{3\theta }{2}\right)-\mathrm{cos}\left(\frac{3\theta }{2}\right)\right]=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{sin}\frac{\theta }{2}=0$ or $\mathrm{sin}\frac{3\theta }{2}-\mathrm{cos}\frac{3\theta }{2}=0$ $⇒\mathrm{sin}\frac{\theta }{2}=\mathrm{sin}0$ or $\mathrm{sin}\frac{3\theta }{2}=\mathrm{cos}\frac{3\theta }{2}$ $⇒$ $\frac{\theta }{2}=n\pi$, $n\in Z$ or $\mathrm{cos}\frac{3\theta }{2}=\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-\frac{3\theta }{2}\right)$ $⇒\theta =2n\pi ,n\in Z$ or $\frac{3\theta }{2}=2m\pi ±\left(\frac{\mathrm{\pi }}{2}-\frac{3\theta }{2}\right),m\in Z$ $⇒$$\theta =2n\pi ,n\in Z$ or $\frac{3\theta }{2}=2m\pi +\frac{\mathrm{\pi }}{2}-\frac{3\theta }{2},m\in Z$ (Taking negative sign will give absurd result.) $\theta =2n\pi ,n\in Z$ or $\theta =\frac{2m\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6},m\in Z$ (vi) $\mathrm{sin}\theta +\mathrm{sin}2\theta +\mathrm{sin}3\theta =0$ $⇒\mathrm{sin}\theta +\mathrm{sin}3\theta +\mathrm{sin}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\left(\frac{4\theta }{2}\right)\mathrm{cos}\left(\frac{2\theta }{2}\right)+\mathrm{sin}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}2\theta \mathrm{cos}\theta +\mathrm{sin}2\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}2\theta \left(2\mathrm{cos}\theta +1\right)=0$ $⇒\mathrm{sin}2\theta =0$ or $2\mathrm{cos}\theta +1=0$ $⇒\mathrm{sin}2\theta =\mathrm{sin}0$ or $\mathrm{cos}\theta =-\frac{1}{2}⇒\mathrm{cos}\theta =\mathrm{cos}\frac{2\mathrm{\pi }}{3}$ $⇒$$\theta =\frac{n\mathrm{\pi }}{2},n\in Z$ or $\theta =2m\pi ±\frac{2\mathrm{\pi }}{3}$, $m\in Z$ (vii) $\mathrm{sin}\theta +\mathrm{sin}2\theta +\mathrm{sin}3\theta +\mathrm{sin}4\theta =0$ $⇒\mathrm{sin}3\theta +\mathrm{sin}\theta +\mathrm{sin}4\theta +\mathrm{sin}2\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\left(\frac{4\theta }{2}\right)\mathrm{cos}\left(\frac{2\theta }{2}\right)+2\mathrm{sin}\left(\frac{6\theta }{2}\right)\mathrm{cos}\left(\frac{2\theta }{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}2\theta \mathrm{cos}\theta +2\mathrm{sin}3\theta \mathrm{cos}\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}\theta \left(\mathrm{sin}2\theta +\mathrm{sin}3\theta \right)=0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}\theta \left(2\mathrm{sin}\left(\frac{5\theta }{2}\right)\mathrm{cos}\left(\frac{\theta }{2}\right)\right)=0\phantom{\rule{0ex}{0ex}}⇒4\mathrm{cos}\theta \mathrm{sin}\left(\frac{5\theta }{2}\right)\mathrm{cos}\left(\frac{\theta }{2}\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{cos}\theta =0,\mathrm{sin}\left(\frac{5\theta }{2}\right)=0$ or $\mathrm{cos}\left(\frac{\theta }{2}\right)=0$ $⇒\mathrm{cos}\theta =\mathrm{cos}\frac{\mathrm{\pi }}{2},\mathrm{sin}\left(\frac{5\theta }{2}\right)=\mathrm{sin}0$ or $\mathrm{cos}\left(\frac{\theta }{2}\right)=\mathrm{cos}\frac{\mathrm{\pi }}{2}$ $⇒\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in Zor\frac{5\theta }{2}=n\pi ,n\in Z$ or, $\frac{\theta }{2}=\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in Z$ $⇒\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in Z$ or $\theta =\frac{2n\mathrm{\pi }}{5},n\in Z$ or $\theta =\left(2n+1\right)\pi ,n\in Z$ (viii) $\mathrm{sin}3\theta -\mathrm{sin}\theta =4{\mathrm{cos}}^{2}\theta -2\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{sin}3\theta -\mathrm{sin}\theta =2\left(2{\mathrm{cos}}^{2}\theta -1\right)\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\left(\frac{2\theta }{2}\right)\mathrm{cos}\left(\frac{4\theta }{2}\right)=2\mathrm{cos}2\theta \phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\theta \mathrm{cos}2\theta =2\mathrm{cos}2\theta \phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta \mathrm{cos}2\theta =\mathrm{cos}2\theta \phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2\theta \left(\mathrm{sin}\theta -1\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{cos}2\theta =0$ or $\mathrm{sin}\theta -1=0$ $⇒$$\mathrm{cos}2\theta =\mathrm{cos}\frac{\mathrm{\pi }}{2}$ or $\mathrm{sin}\theta =1⇒\mathrm{sin}\theta =\mathrm{sin}\frac{\mathrm{\pi }}{2}$ $⇒2\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2}$, $n\in Z$ or $\theta =n\pi +\left(-1{\right)}^{n}\frac{\mathrm{\pi }}{2},n\in Z$ $⇒\theta =\left(2n +1\right)\frac{\mathrm{\pi }}{4},n\in Z$ or $\theta =n\pi +\left(-1{\right)}^{n}\frac{\mathrm{\pi }}{2},n\in Z$ (ix) $\mathrm{sin}2\theta -\mathrm{sin}4\theta +\mathrm{sin}6\theta =0.$ $⇒2\mathrm{sin}\left(\frac{8\theta }{2}\right)\mathrm{cos}\left(\frac{4\theta }{2}\right)-\mathrm{sin}4\theta =0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}4\theta \mathrm{cos}2\theta -\mathrm{sin}4\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}4\theta \left(2\mathrm{cos}2\theta -1\right)=0\phantom{\rule{0ex}{0ex}}$ $⇒\mathrm{sin}4\theta =0$ or $2\mathrm{cos}2\theta -1=0$ $⇒4\theta =n\pi$, $n\in Z$ or $\mathrm{cos}2\theta =\frac{1}{2}⇒\mathrm{cos}2\theta =\mathrm{cos}\frac{\mathrm{\pi }}{3}$ $⇒\theta =\frac{n\mathrm{\pi }}{4},n\in Z$ or $\theta =n\pi ±\frac{\mathrm{\pi }}{6},n\in Z$

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