Question

7cos²x+3sin²x=4(find general solution)​

Answer 1

Answer:

The solution for x is

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

Step-by-step explanation:

Given : Expression $7\cos^2x+3\sin^2x=47cos$

To find : Solve the given expression ?

Solution:-

We solve the expression by splitting the term,

7$\cos^2x+3\sin^2x=47cos$

4$\cos^2x+3\cos^2x+3\sin^2x=44cos$

4$\cos^2x+3(\cos^2x+\sin^2x)=44cos$

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

4$\cos^2x+3(1)=44cos$

4$\cos^2x=14cos$

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

$\cos x=\pm\frac{1}{2}cosx=±$

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

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

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

II case : $\cos x=-\frac{1}{2}cosx=−$

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

$\begin{gathered}x =\pi+\frac{\pi}{3} \\\\x =\frac{4\pi }{3}\end{gathered}$

Therefore, The solution for x is

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

Answer 2

Answer:

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