Question

$$\begin{array}{cc}\text{Trigonometric Equation}& \text{Principal Solutions}\\ 1.\sin x=\frac{\sqrt{3}}{2}& A.\frac{5\pi }{6}\\ 2.\tan x=\frac{-1}{\sqrt{3}}& B.\frac{5\pi }{12}\\ 3.\sec 2x=\frac{-2}{\sqrt{3}}& C.\frac{7\pi }{12}\\ 4.\cos 3x=\left(\frac{-1}{2}\right)& D.\frac{\pi }{3}\\ & E.\frac{2\pi }{9}\\ & F.\frac{4\pi }{9}\\ & G.\frac{11\pi }{6}\\ & H.\frac{2\pi }{3}\end{array}$$

• A. 1 - D, 2 - A, 3 - B, 4 - E
• B. 1 - D, 2 - G, 3 - B, 4 - F
• C. 1 - H, 2 - A, 3 - C, 4 - F
• D. 1 - H, 2 - G, 3 - C, 4 - E

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

1 - D, 2 - A, 3 - B, 4 - E

B

1 - D, 2 - G, 3 - B, 4 - F

C

1 - H, 2 - A, 3 - C, 4 - F

D

1 - H, 2 - G, 3 - C, 4 - E

We have learnt that the values of $\sin x$ & $\cos x$ repeat after an interval of $2\pi$ and the values of $\tan x$ repeat after an interval of $pi$. The solutions of trigonometric equations for which $x$ lies between 0 and $2\pi$ are called principal solutions.

So,

$1.\sin x=\frac{\sqrt{3}}{2}$

We know $\sin x$ is positive in $1^{st}$ & $2^{nd}$ quadrant in the interval of 0 to 2$\pi$.

in $1^{st}$ quadrant, $\sin \frac{\pi }{3}=\frac{\left(\sqrt{3}\right)}{2}$

in $2^{nd}$ quadrant $\sin (\pi -\frac{\pi }{3})=\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$

Therefore, principal solution of equation $\sin x=\frac{\sqrt{3}}{2}\text{are}\frac{\pi }{3},\frac{2\pi }{3}$

$2.\tan x=-\frac{1}{\sqrt{3}}$

$\tan x$ is negative in $2^{nd}$ & $4^{th}$ quadrant in the interval of 0 to $2\pi$.

$\text{We know that}$ $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$. $\tan (\pi -\frac{\pi }{6})=-\tan \frac{\pi }{6}=-\frac{1}{\sqrt{3}}$

So, $\pi -\frac{\pi }{6}=\frac{5\pi }{6}$ and

in $4^{th}$ quadrant $(2\pi -\frac{\pi }{6})=-\tan \frac{\pi }{6}=-\frac{1}{\sqrt{3}}$
$(2\pi -\frac{\pi }{6})=\frac{11\pi }{6}$

Therefore, Principal solutions are $\frac{5\pi }{6}\text{and}\frac{11\pi }{6}$

$3.\sec 2x=-\frac{2}{\sqrt{3}}$
$\frac{1}{\cos 2x}=-\frac{2}{\sqrt{3}}$
$\cos 2x=-\frac{\sqrt{3}}{2}$

$\cos \theta$ is negative in $2^{nd}$ & $3^{rd}$ quadrant in the interval of 0 to $2\pi$.

We know, $\cos \frac{\pi }{6}$ = $\frac{\sqrt{3}}{2}$

in $2^{nd}$ quadrant $\cos (\pi -\frac{\pi }{6})=-\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$

Also, $\pi -\frac{\pi }{6}=\frac{5\pi }{6}$
$\Rightarrow 2x=\frac{5\pi }{6}$
or $x=\frac{5\pi }{12}$

In $3^{rd}$ quadrant $\cos (\pi +\frac{\pi }{6})=-\cos \frac{\pi }{6}=-\frac{\sqrt{3}}{2}$
$\Rightarrow 2x=\pi +\frac{\pi }{6}=\frac{7\pi }{6}$
or $x=\frac{7\pi }{12}$

Therefore, principal solutions of equation $\sec 2x=-\frac{2}{\sqrt{3}}$ are $\frac{5\pi }{12}\text{and}\frac{7\pi }{12}$
4. $\cos 3x=-\frac{1}{2}$

$\cos \theta$ is negative in $2^{nd}$ & $3^{rd}$ quadrant

$\cos \frac{\pi }{3}=\frac{1}{2}$

In $2^{nd}$ quadrant $\cos (\pi -\frac{\pi }{3})=-\cos \frac{\pi }{3}=-\frac{1}{2}$
$\Rightarrow 3x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
$x=\frac{2\pi }{9}$

In $3^{rd}$ quadrant $\cos (\pi +\frac{\pi }{3})=-\cos \frac{\pi }{3}=-\frac{1}{2}$
$3x=\pi +\frac{\pi }{3}=\frac{4\pi }{3}$
$x=\frac{4\pi }{9}$

Therefore, principal solutions of equation $\cos 3x=-\frac{1}{2}$ are $\frac{2\pi }{9}$ and $\frac{4\pi }{9}$.