Question

determine the value of x such that 2 cosec square 30 degree + x sin square 60 degree minus 3/4 tan square 30 degree = 10

Answer 1

hope it is correct .....
2*4+x*3/4-3/4*1/3=10
8+3x/4-3/4*1/3=10
3x/4-3/4*1/3=10-8
3x/4-1/4=2
(3x-1)=2*4
3x-1=8
3x=8+1
x=9/3=>3

Answer 2

Answer:

The value of x is 3.

Step-by-step explanation:

Given,

$2cosec^2 30^{\circ}+x sin^2 60^{\circ} - \frac{3}{4} tan^2 30^{\circ}=10$

$2 (2)^2 + x ( \frac{\sqrt{3}}{2})^2 - \frac{3}{4}\times (\frac{1}{\sqrt{3}})^2 = 10$

$8 + \frac{3x}{4}-\frac{3}{12}=10$

$\frac{3x}{4}=10+\frac{1}{4}-8$

$\frac{3x}{4}=\frac{40+1-32}{4}$

$\frac{3x}{4}=\frac{9}{4}$

$\implies x = \frac{36}{12}= 3$

Thus, the value of x is 3.