Question

$$\begin{array}{cc}\text{Trigonometric Equations}& \text{General Solutions}\\ 1.7{\cos }^{2}x+3{\sin }^{2}x=4& P.n\pi \pm \frac{\pi }{4},\text{where}n\in I\\ 2.{\sin }^{2}x=\frac{1}{2}& Q.n\pi \pm \frac{\pi }{3},\text{where}n\in I\\ 3.{\tan }^{2}x+3=0& R.n\pi \pm \frac{\pi }{6},\text{where}n\in I\\ 4.3{\tan }^{2}x-1=0& S.\text{No solution}\end{array}$$

Answer 1

1. $7co{s}^{2}x+3{\sin }^{2}x=4$

We know, identity ${\sin }^{2}x+{\cos }^{2}x=1$

$\Rightarrow (4+3){\cos }^{2}x+3{\sin }^{2}x=4$

$\Rightarrow 4{\cos }^{2}x+3{\cos }^2+3{\sin }^{2}x=4$

$\Rightarrow 4{\cos }^{2}x+3({\sin }^{2}x+{\cos }^{2}x)=4$

$\Rightarrow 4{\cos }^{2}x+3=4$

$\Rightarrow 4{\cos }^{2}x=1$

$\Rightarrow {\cos }^{2}x=\frac{1}{4}$

$\Rightarrow {\cos }^{2}x={(\frac{1}{2})}^2={\cos }^2\frac{\pi }{3}$

General solutions is $x=n\pi \pm \frac{\pi }{3}$ , Where n ∈ Z

2 . ${\sin }^{2}x=\frac{1}{2}$

${\sin }^{2}x={(\frac{1}{\sqrt{3}})}^2={\sin }^2\frac{\pi }{4}$

General solution is $n\pi \pm \frac{\pi }{4}$

3. ${\tan }^{2}x+3=0$

${\tan }^{2}x=-3$

Square of trigonometric function cannot be negative .

No solution

4. $3{\tan }^{2}x-1=0$

${\tan }^{2}x=\frac{1}{3}$

${\tan }^{2}x=(\frac{1}{\sqrt{3}}{)}^2$

${\tan }^{2}x={\tan }^2\frac{\pi }{6}$

So, general solution is $x=n\pi \pm \frac{\pi }{6}$

Hence the correct answer is Option a.