Question

Please answer 11 question?

Class - 10th
Chapter - Trigonometry

I swear spammed answer ID report .

Answer 1

Given:-

$\sqrt{3} sin \: x - cos \: x = 0$

To find:-

Value of x

Solution:-

$\sqrt{3} sin \: x - cos \: x = 0$

$= > \sqrt{3} sin \: x = cos \: x$

$= > \sqrt{3} sin \: x = \sqrt{1 - {sin}^{2}x }$

$= > { (\sqrt{3}sin \: x) }^{2} = 1 - {sin}^{2} x$

$= > 3 {sin}^{2}x = 1 - {sin}^{2} x$

$= > 4 {sin}^{2} x = 1$

$= > {sin}^{2} x = \frac{1}{4}$

$= > sin \: x = \sqrt{ \frac{1}{4} }$

$= > sin \: x = \frac{1}{2}$

We know that,

sin 30° = 1/2 -(1)

but, sin x = 1/2 -(2)

From (1) and (2), we get,

x = 30°

Answer 2

${\huge{\underline{\underline{\bf{\maltese Question}}}}}$

Q - If √3 sin x - cos x = 0 and 0°<x<90° , find the value of x .

${\huge{\underline{\underline{\bf{\maltese Answer\checkmark}}}}}$

In this question ,

All we have to do is just Modifying the Given equation .

$\rm \: \sqrt{3} \: sin \: x - cos \: x = 0$

$\rm \implies \: \sqrt{3} \: sin \: x = cos \: x$

$\implies \: \rm \: \dfrac{ \sqrt{3} \: sin \: x }{cos \: x} = 1$

$\implies \rm \: \frac{sin \: x}{cos \: x} = \frac{1}{ \sqrt{3} }$

$\rm\implies \: tan \: x = \frac{1}{ \sqrt{3} }$

Now ,

Using the trigonometric table ,

$\rm \: tan \: x \: = \: tan \: 30 \degree$

So , on comparing ,

$\pink\bigstar\boxed{ \rm \: x = 30 \degree}$

___________

★ Know More :

• $\sf \: tan \: \theta = \frac{sin \: \theta}{cos \: \theta}$
• $\sf \: sec \: \theta = \frac{1}{cos \: \theta}$
• $\sf \: cosec \: \theta = \frac{1}{sin \: \theta}$
• $\sf \: {sin}^{2} \: x + {cos}^{2} \: x = 1$
• $\sf \: 1 + {cot}^{2} \: x = {cosec}^{2} \: x$
• $\sf \: {tan}^{2} \: x + 1 = {sec}^{2} \: x$