Question

"$$\begin{array}{cc}\text{Trigonometric Equation}& \text{Principal Solutions}\\ 1.sinx=\frac{\sqrt{3}}{2}& A.\frac{5\pi }{6}\\ 2.tanx=\frac{-1}{\sqrt{3}}& B.\frac{5\pi }{12}\\ 3.sec2x=\frac{-2}{\sqrt{3}}& C.\frac{7\pi }{12}\\ 4.cos3x=\left(\frac{-1}{2}\right)& D.\frac{\pi }{3}\\ & E.\frac{2\pi }{9}\\ & F.\frac{4\pi }{9}\\ & G.\frac{11\pi }{6}\\ & H.\frac{2\pi }{3}\end{array}$$ "

• 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 - j
• 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 - j

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,

$1.sinx=\frac{\sqrt{3}}{2}$

We know sin x is positive in i & ii quadrant in the interval of 0 to 2 $\pi$.

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

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

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

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

tan x is negative in ii & iv 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 iv 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 $\frac{5\pi }{6}and\frac{11\pi }{6}$

$3.sec2xc=-\frac{2}{\sqrt{3}}$
$\frac{1}{cos2x}=-\frac{2}{\sqrt{3}}$
$cos2x=-\frac{\sqrt{3}}{2}$

$Cos\pi$ is negative in ii & iii quadrant in the interval of 0 to $2\pi$.

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

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

$\pi -\frac{\pi }{6}=\frac{5\pi }{6}$
$2x=\frac{5\pi }{6}$
$x=\frac{5\pi }{12}$

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

Therefore, principal solutions of equations $sec2x=-\frac{2}{\sqrt{3}}$ are $\frac{5\pi }{12}$ & $\frac{7\pi }{2}$
$cos3x=-\frac{1}{2}$

$cos\theta$ is negative in ii & iii quadrant

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

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

In iii 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 $cos3x=-\frac{1}{2}$ are $\frac{2\pi }{9}$ & $\frac{4\pi }{9}$