Question

Solve 7cos2x + 3sin2x = 4

Answer 1

.......................

Answer 2

Answer:

The solution for x is

$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$

Step-by-step explanation:

Given : Expression $7\cos^2x+3\sin^2x=4$

To find : Solve the given expression ?

Solution :

We solve the expression by splitting the term,

$7\cos^2x+3\sin^2x=4$

$4\cos^2x+3\cos^2x+3\sin^2x=4$

$4\cos^2x+3(\cos^2x+\sin^2x)=4$

We know, $\cos^2x+\sin^2x=1$

$4\cos^2x+3(1)=4$

$4\cos^2x=1$

$\cos^2x=\frac{1}{4}$

$\cos x=\pm\frac{1}{2}$

I case : $\cos x=\frac{1}{2}$

$x=\frac{\pi}{3}$

$x = 2\pi-\frac{\pi}{3} \\\\x =\frac{5\pi }{3}$

II case : $\cos x=-\frac{1}{2}$

$x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

$x =\pi+\frac{\pi}{3} \\\\x =\frac{4\pi }{3}$

Therefore, The solution for x is

$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$