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Question Number 70597 by Maclaurin Stickker last updated on 06/Oct/19

$$solve$$$$\cos {}^2\beta +\cos {}^{2}3\beta =1$$

Answered by Kunal12588 last updated on 08/Oct/19

$$cos2\alpha =2{cos}^2\alpha -1\Rightarrow {cos}^2\alpha =\frac{1+cos2\alpha }{2}$$$${cos}^2\beta =\frac{1+cos2\beta }{2}$$$${cos}^{2}3\beta =\frac{1+cos6\beta }{2}$$$$\therefore 1+cos2\beta +1+cos6\beta =2$$$$\Rightarrow cos2\beta +cos6\beta =0$$$$\Rightarrow 2cos4\beta cos2\beta =0$$$$\Rightarrow cos4\beta =0andcos2\beta =0$$$$\Rightarrow 4\beta =2n\pi \pm \frac{\pi }{2}and2\beta =2n\pi \pm \frac{\pi }{2}$$$$\Rightarrow \beta =\frac{n\pi }{2}\pm \frac{\pi }{8},n\pi \pm \frac{\pi }{4}$$

Commented by mr W last updated on 06/Oct/19

$$niceway!$$$$bettersolutionthanmine!$$

Commented by mathmax by abdo last updated on 06/Oct/19

$$error{cos}^2\alpha =\frac{1+cos\left(2\alpha \right)}{2}..$$

Commented by Kunal12588 last updated on 08/Oct/19

$$thankssir$$

Answered by mr W last updated on 06/Oct/19

$$\cos 3\beta =\cos \beta (4{\cos }^2\beta -3)$$$${\cos }^2\beta +{\cos }^2\beta {(4{\cos }^2\beta -3)}^2=1$$$$lett={\cos }^2\beta ⩾0,⩽1$$$$\Rightarrow t+t{(4t-3)}^2=1$$$$\Rightarrow 16t^3-24t^2+10t-1=0$$$$lets=2t⩾0,⩽2$$$$\Rightarrow 2s^3-6s^2+5s-1=0$$$$\Rightarrow (s-1)(2s^2-4s+1)=0$$$$\Rightarrow s=1$$$$\Rightarrow s=\frac{2\pm \sqrt{2}}{2}$$$$\Rightarrow t={\cos }^2\beta =\frac{1}{2}\Rightarrow \cos \beta =\pm \frac{\sqrt{2}}{2}\Rightarrow \beta =n\pi \pm \frac{\pi }{4}$$$$\Rightarrow t={\cos }^2\beta =\frac{2+\sqrt{2}}{4}\Rightarrow \cos \beta =\pm \frac{\sqrt{2+\sqrt{2}}}{2}\Rightarrow \beta =n\pi \pm \frac{\pi }{8}$$$$\Rightarrow t={\cos }^2\beta =\frac{2-\sqrt{2}}{4}\Rightarrow \cos \beta =\pm \frac{\sqrt{2-\sqrt{2}}}{2}\Rightarrow \beta =n\pi \pm \frac{3\pi }{8}$$

Answered by Rasheed.Sindhi last updated on 06/Oct/19

$$\cos {}^2\beta +\cos {}^{2}3\beta =1$$$$\cos {}^2\beta +\cos {}^{2}3\beta ={\sin }^2\beta +{\cos }^2\beta$$$$\cos {}^{2}3\beta ={\sin }^2\beta$$$$\cos {}^{2}3\beta -{\sin }^2\beta =0$$$$(\cos 3\beta -\sin \beta )(\cos 3\beta +\sin \beta )=0$$$$\cos 3\beta -\sin \beta =0\vee \cos 3\beta +\sin \beta )$$$$\cos 3\beta =\sin \beta \vee \cos 3\beta =-\sin \beta )$$$$\cos 3\beta =\sin \beta \vee \cos 3\beta =\sin (-\beta )$$$$3\beta +\beta =\pi /2\vee 3\beta -\beta =\pi /2$$$$4\beta =\frac{\pi }{2}\vee 2\beta =\frac{\pi }{2}$$$$\beta =\frac{\pi }{8},\frac{\pi }{4}$$

Answered by ajfour last updated on 06/Oct/19

$$2\beta =\theta$$$$\cos {}^2(\theta -\beta )+\cos {}^2(\theta +\beta )=1$$$$\cos (2\theta -2\beta )+\cos (2\theta +2\beta )=0$$$$\Rightarrow 2\theta +2\beta =(2n+1)\pi \pm (2\theta -2\beta )$$$$\Rightarrow 6\beta =(2n+1)\pi \pm 2\beta$$$$\Rightarrow \beta =\frac{(2n+1)\pi }{8},\beta =\frac{(2n+1)\pi }{4}$$$$\Rightarrow \beta =\frac{k\pi }{8}.$$

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