Damion Ellis

2023-02-23

The number of solutions of $\mathrm{cos}\left(2\theta \right)=\mathrm{sin}\left(\theta \right)$ in $\left(0,2\pi \right)$ is...

Roman Hogan

$⇒1-2{\mathrm{sin}}^{2}\left(\theta \right)=\mathrm{sin}\left(\theta \right)\left(\because \mathrm{cos}\left(2x\right)={\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)&{\mathrm{cos}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(x\right)=1\right)⇒2{\mathrm{sin}}^{2}\left(\theta \right)+\mathrm{sin}\left(\theta \right)-1=0⇒2{\mathrm{sin}}^{2}\left(\theta \right)+2\mathrm{sin}\left(\theta \right)-\mathrm{sin}\left(\theta \right)-1=0⇒2\mathrm{sin}\left(\theta \right)\left(\mathrm{sin}\left(\theta \right)+1\right)-\left(\mathrm{sin}\left(\theta \right)+1\right)=0⇒\left(\mathrm{sin}\left(\theta \right)+1\right)\left(2\mathrm{sin}\left(\theta \right)-1\right)=0$
$\therefore \left(\mathrm{sin}\left(\theta \right)+1\right)=0$ or $\left(2\mathrm{sin}\left(\theta \right)-1\right)=0$
$\therefore \mathrm{sin}\left(\theta \right)=-1$ or $\mathrm{sin}\left(\theta \right)=\frac{1}{2}$
$\because \theta \in \left(0,2\pi \right),\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$

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