Question

The general solution of the equation $7{\cos }^2\mathrm{\theta }+3{\sin }^2\mathrm{\theta }=4$ is
(a) $\mathrm{\theta }=2n\pi \pm \frac{\pi }{6},n\in Z$

(b) $\mathrm{\theta }=2n\pi \pm \frac{2\pi }{3},n\in Z$

(c) $\mathrm{\theta }=n\pi \pm \frac{\pi }{3},n\in Z$

(d) none of these

Answer 1

(c) $\mathrm{\theta }=n\pi \pm \frac{\pi }{3},n\in Z$
Given:

$$\begin{gathered} 7{\cos }^2\theta +3{\sin }^2\theta =4 \\ \Rightarrow 7{\cos }^2\theta +3(1-{\cos }^2\theta )=4 \\ \Rightarrow 7{\cos }^2\theta +3-3{\cos }^2\theta =4 \\ \Rightarrow 4{\cos }^2\theta +3=4 \\ \Rightarrow 4(1-{\cos }^2\theta )=3 \\ \Rightarrow 4{\sin }^2\theta =3 \\ \Rightarrow {\sin }^2\theta =\frac{3}{4} \\ \Rightarrow \sin \theta =\frac{\sqrt{3}}{2} \\ \Rightarrow \sin \theta =\sin \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z} \end{gathered}$$