Question

If $\sqrt{3}\left(\cos x{)}^2=\right(\sqrt{3}-1)\cos x+1$ the number of solutions of the given equation when $x\in \left[0,\frac{\mathrm{\pi }}{2}\right]$ is:

Answer 1

Solve the equation:

Step 1: Factorize the equation

Given that, $\sqrt{3}(\cos x{)}^2=(\sqrt{3}-1)\cos x+1$ first, we factorize the taking common factor.

$$\begin{gathered} \Rightarrow \sqrt{3}{\left(\cos x\right)}^2=\sqrt{3}\mathrm{co}sx-\cos x+1 \\ \Rightarrow \sqrt{3}(\cos x{)}^2-\sqrt{3}\cos x+\cos x-1=0 \\ \Rightarrow \sqrt{3}\cos x(\cos x-1)+1(\mathrm{co}sx-1)=0 \\ \Rightarrow (\sqrt{3}\cos x+1)(\cos x-1)=0 \end{gathered}$$
Step 2: Using the zero property of the product we have:
$\Rightarrow \cos x=1\text{or}\cos x=-\frac{1}{\sqrt{3}}$
But $\cos x=-\frac{1}{\sqrt{3}}$ not possible.
And $\cos x=1$ gives $x=0$.

Hence, the number of solutions is one.