Question

Find general solution (√3+1)cos x + (√3-1) sin x=2

Answer 1

Hope this answer helps you.

Answer 2

Answer:

$x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$

Step-by-step explanation:

Given : Expression  $(\sqrt3+1)\cos x+(\sqrt3-1)\sin x=2$

To find : The general solution of the expression ?

Solution :

Let $(\sqrt3+1)=r\cos \alpha$

$(\sqrt3-1)=r\sin\alpha$

Substitute in expression,

$r\cos \alpha\cos x+r\sin\alpha\sin x=2$

$r(\cos \alpha\cos x+\sin\alpha\sin x)=2$

$r(\cos(x-\alpha)=2$

$\cos(x-\alpha=\frac{2}{r}$

We know,

$r=\sqrt{(\sqrt3-1)^2+(\sqrt3+1)^2}\\r=\sqrt{4-2\sqrt3+4+2\sqrt3}\\r=\sqrt8\\r=2\sqrt2$

$\cos(x-\alpha=\frac{2}{2\sqrt2}$

$\cos(x-\alpha=\frac{1}{\sqrt2}$

$\cos(x-\alpha=\cos(\frac{\pi}{4})$

$(x-\alpha)=2n\pi\pm\frac{\pi}{4}$

Now, $\tan\alpha=\frac{\sqrt3-1}{\sqrt3+1}$

$\tan\alpha=\tan(\frac{\pi}{3}-\frac{\pi}{4}$

$\alpha=\frac{\pi}{12}$

Substitute,

$(x-\alpha)=2n\pi\pm\frac{\pi}{4}$

$x-\frac{\pi}{12}=2n\pi\pm\frac{\pi}{4}$

$x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$

Therefore, The general solution is $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$