Question

Determine the value of x such that 2cosec^2 30°+ xsin^2 60°- 3/4tan^3 30°= 10​

Answer 1

Step-by-step explanation:

do the next step yourself.

Answer 2

Given:

$\sf{2cosec^{2}30^{\circ}+xsin^{2}60^{\circ}-\dfrac{3}{4}tan^{3}30^{\circ}=10}$

To Find:

• Value of x

Solution:

We know that,

$\longrightarrow\bf{cosec30^{\circ}=2}$

$\longrightarrow\bf{sin60^{\circ}=\dfrac{\sqrt{3}}{2}}$

$\longrightarrow\bf{tan30^{\circ}=\dfrac{1}{\sqrt{3}}}$

Now,

$\sf{2cosec^{2}30^{\circ}+xsin^{2}60^{\circ}-\dfrac{3}{4}tan^{3}30^{\circ}=10}$

$\sf{2(2)^{2}+x\bigg(\dfrac{\sqrt{3}}{2}\bigg)^{2}-\dfrac{3}{4}\bigg(\dfrac{1}{\sqrt{3}}\bigg)^{3}=10}$

$\sf{2(4)+\dfrac{3x}{4}-\dfrac{3}{4}\bigg(\dfrac{1}{3\sqrt{3}}\bigg)=10}$

$\sf{8+\dfrac{3x}{4}-\dfrac{1}{4\sqrt{3}}=10}$

$\sf{\dfrac{32\sqrt{3}+3\sqrt{3}x-1}{4\sqrt{3}}=10}$

$\sf{32\sqrt{3}+3\sqrt{3}x-1=40\sqrt{3}}$

$\sf{3\sqrt{3}x=40\sqrt{3}-32\sqrt{3}+1}$

$\sf{3\sqrt{3}x=8\sqrt{3}+1}$

$\sf\pink{x=\dfrac{8\sqrt{3}+1}{3\sqrt{3}}}$

Hence the value of x is $\sf\green{\dfrac{8\sqrt{3}+1}{3\sqrt{3}}}$