Question

Find the value of $xϵ(0,\pi )$ which satisfies the equation sin x +$\sqrt{3}$ cos x = $\sqrt{2}$

• A. $\frac{\pi }{12}$
• B. $\frac{5\pi }{12}$
• C. $\frac{11\pi }{12}$
• D. $\frac{7\pi }{12}$

Answer 1

The correct option is B

$\frac{5\pi }{12}$

When we get equations of the form a $sin\theta +bcos\theta =c$, we put a = r cos A and b = r sin A. This is similar

to the method we follow while finding the maximum value of a sin θ + b cosθ
We divide and multiply by $\sqrt{a^2+b^2}$ to get terms like cos $(A\mp B)$ or $sin(A\mp B)$ on one side.

The given expression is
Sin x + $\sqrt{3}$ cos x = $\sqrt{2}$
$\Rightarrow$ we will divide and multiply by $\sqrt{(\sqrt{3}{)}^2+1}$ = 2
$\Rightarrow$ $2(sinx\times \frac{1}{2}+\frac{\sqrt{3}}{2}cosx)=\sqrt{2}$
Now we will replace $\frac{1}{2}and\frac{\sqrt{3}}{2}$ by cos A or sin A.
$\Rightarrow$ sin x $\times$ sin $\frac{\pi }{6}+cos\frac{\pi }{6}cosx=\frac{\sqrt{2}}{2}$
$\Rightarrow$ cos $(x-\frac{\pi }{6})=cos\frac{\pi }{4}$
$\Rightarrow$ x - $\frac{\pi }{6}=2n\pi \mp \frac{\pi }{4}$
We will split the expressions into $x=2n\pi -\frac{\pi }{4}+\frac{\pi }{6}orx=2n\pi +\frac{5\pi }{12}$
The value of x which lie in $(0,\pi )is\frac{5\pi }{12}$ (we get this by putting n = 1, 2 ............)