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Question

# The general solution of the equation $7{\mathrm{cos}}^{2}\mathrm{\theta }+3{\mathrm{sin}}^{2}\mathrm{\theta }=4$ is (a) $\mathrm{\theta }=2n\pi ±\frac{\pi }{6},n\in Z$ (b) $\mathrm{\theta }=2n\pi ±\frac{2\pi }{3},n\in Z$ (c) $\mathrm{\theta }=n\pi ±\frac{\pi }{3},n\in Z$ (d) none of these

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Solution

## (c) $\mathrm{\theta }=n\pi ±\frac{\pi }{3},n\in Z$ Given: $7{\mathrm{cos}}^{2}\theta +3{\mathrm{sin}}^{2}\theta =4\phantom{\rule{0ex}{0ex}}⇒7{\mathrm{cos}}^{2}\theta +3\left(1-{\mathrm{cos}}^{2}\theta \right)=4\phantom{\rule{0ex}{0ex}}⇒7{\mathrm{cos}}^{2}\theta +3-3{\mathrm{cos}}^{2}\theta =4\phantom{\rule{0ex}{0ex}}⇒4{\mathrm{cos}}^{2}\theta +3=4\phantom{\rule{0ex}{0ex}}⇒4\left(1-{\mathrm{cos}}^{2}\theta \right)=3\phantom{\rule{0ex}{0ex}}⇒4{\mathrm{sin}}^{2}\theta =3\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta =\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\mathrm{sin}\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒\theta =n\mathrm{\pi }±\frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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