Question

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

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

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

(d) none of these

Answer 1

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

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