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Question Number 29354 by NECx last updated on 08/Feb/18

$$3\sin {}^2\theta -\sin \theta \cos \theta -4\cos {}^2\theta =0$$$$findthevaluesof\theta if\theta lies$$$$between0and360$$

Answered by mrW2 last updated on 08/Feb/18

$$3(1-{\cos }^2\theta )-\sin \theta \cos \theta -4\cos {}^2\theta =0$$$$3-\sin \theta \cos \theta -7\cos {}^2\theta =0$$$$6-2\sin \theta \cos \theta -7\left(2\cos {}^2\theta \right)=0$$$$6-\sin 2\theta -7(1+\cos 2\theta )=0$$$$\sin 2\theta +7\cos 2\theta =-1$$$$\sin 2\theta \times \frac{1}{\sqrt{7^2+1^1}}+\cos 2\theta \times \frac{7}{\sqrt{7^2+1^2}}=-\frac{1}{\sqrt{7^2+1^2}}$$$$\sin 2\theta \times \sin \alpha +\cos 2\theta \times \cos \alpha =-\frac{1}{5\sqrt{2}}$$$$with\alpha ={\tan }^{-1}\frac{1}{7}$$$$\Rightarrow \cos (2\theta -\alpha )=-\frac{1}{5\sqrt{2}}$$$$\Rightarrow 2\theta -\alpha =(2n+1)\pi \pm {\cos }^{-1}\frac{1}{5\sqrt{2}}$$$$\Rightarrow \theta =n\pi +\frac{1}{2}(\pi +\alpha \pm {\cos }^{-1}\frac{1}{5\sqrt{2}})$$$$\Rightarrow \theta =n\pi +\frac{1}{2}(\pi +{\tan }^{-1}\frac{1}{7}\pm {\cos }^{-1}\frac{1}{5\sqrt{2}})$$$$\Rightarrow \theta =180n+\{\begin{array}{l}135°\\ 53.1°\end{array}$$$$solutionswithin0°-360°:$$$$\Rightarrow \theta =53.1°,135°,233.1°,315°$$

Answered by mrW2 last updated on 08/Feb/18

$$anotherway:$$$$(3\sin \theta -4\cos \theta )(\sin \theta +\cos \theta )=0$$$$\Rightarrow 3\sin \theta -4\cos \theta =0$$$$\Rightarrow \sin \theta +\cos \theta =0$$$$\Rightarrow \tan \theta =\frac{4}{3}$$$$\Rightarrow \tan \theta =-1$$$$\Rightarrow \theta =n\pi +{\tan }^{-1}\frac{4}{3}or180n+53.1°$$$$\Rightarrow \theta =n\pi -\frac{\pi }{4}or180n-45°$$$$$$$$within0-360°:$$$$\theta =53.1°,135°,233.1°,315°$$

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