Solve the equation: $2{{\sin }^{2}}x+\sqrt{3}\cos x+1=0$

Hint: To solve this equation we will use some trigonometric formulas to convert sin into cos and then we can see that it has been converted into quadratic equation, so now we are going to use Sridharacharya formula to find the roots of the equation and from that we can find the value of x.

Complete step-by-step answer:

Let’s first write the trigonometric formula that we will use,

${{\sin }^{2}}x=1-{{\cos }^{2}}x$

Now we will use sridharacharya formula to find the roots of this quadratic equation,

The formula is:

If the equation is $a{{x}^{2}}+bx+c=0$ , then the formula for finding x is:

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

This equation $2{{\sin }^{2}}x+\sqrt{3}\cos x+1$ becomes,

$\begin{align}

  & 2\left( 1-{{\cos }^{2}}x \right)+\sqrt{3}\cos x+1=0 \\

 & 2-2{{\cos }^{2}}x+\sqrt{3}\cos x+1=0 \\

 & 2{{\cos }^{2}}x-\sqrt{3}\cos x-3=0 \\

\end{align}$

Now we are going to use sridharacharya formula,

$\begin{align}

  & \cos x=\dfrac{-\left( -\sqrt{3} \right)\pm \sqrt{{{\left( -\sqrt{3} \right)}^{2}}-4\left( 2 \right)\left( -3 \right)}}{2\times 2} \\

 & \cos x=\dfrac{\sqrt{3}\pm \sqrt{3+24}}{4} \\

 & \cos x=\dfrac{\sqrt{3}\pm 3\sqrt{3}}{4} \\

\end{align}$

Now from this we get two values of cosx but one is wrong,

Let’s see which one is wrong and how,

$\cos x=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\text{ and }\cos x=\dfrac{-2\sqrt{3}}{4}=\dfrac{-\sqrt{3}}{2}$

We know that the range of cosx is from -1 to +1 and hence the value of $\sqrt{3}$ does not lie in the range and hence it is the wrong answer.

Now the second value of cosx is correct,

And hence it is the correct answer.

Therefore,

$\begin{align}

  & \cos x=\dfrac{-\sqrt{3}}{2} \\

 & \cos x=\cos \dfrac{5\pi }{6} \\

\end{align}$

If $\cos x=\cos \theta $ ,

Then the formula for the general solution of x is $2n\pi \pm \theta $
.
Here $\theta =\dfrac{5\pi }{6}$ hence, the general solution of x is $2n\pi \pm \dfrac{5\pi }{6}$ , where n is an integer.

Note: In this question students might get confused that the given equations are in trigonometric form but we have converted it into quadratic of cos and then solved it. Another method to solve this question is to convert it into the quadratic of sin and then find the solution.