Question

please answer the question. ....​

Answer 1

We are given a Trigonometric Equation to solve.

In this case, we have a $\cos^2x$ and a $\sin x$. It suggests us that we might be able to form a quadratic equation in $\sin x$.

For that, we will use one simple identity:

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

And so here we can solve as follows:

$\mathsf{2\cos^2x-5\sin x+1=0} \\\\\\ \mathsf{\implies 2(1-\sin^2x)-5\sin x+1=0} \\\\\\ \mathsf{\implies 2-2\sin^2x-5\sin x+1=0}\\\\\\\mathsf{\implies 2\sin^2x+5\sin x-3=0}\\\\\\\textsf{Now we split the middle term}\\\\\\\mathsf{\implies 2\sin^2x+6\sin x-\sin x-3=0}\\\\\\\mathsf{\implies 2\sin x(\sin x+3)-1(\sin x+3)=0}\\\\\\\mathsf{\implies (\sin x+3)(2\sin x-1)=0}\\\\\\\mathsf{\implies \sin x+3=0 \quad OR \quad 2\sin x -1=0}\\\\\\\mathsf{\implies \sin x=-3 \quad OR \quad \sin x=\dfrac{1}{2}}$

$\textsf{But sin x = -3 is not possible.}$

Now, if you haven't studied General Solutions of Trigonometric Equations, then your answer will be:

$\mathsf{\implies \sin x=\dfrac{1}{2}} \\ \\ \\ \implies \huge \boxed{\bold{x=30^{\circ}}}$

However, if you do know General Solutions of Trigonometric Equations, then your answer will be:

$\mathsf{\sin x=\dfrac{1}{2}}\\\\\\\mathsf{\implies \sin x=\sin \dfrac{\pi}{6}}\\\\\\ \implies \huge \boxed{\bold{x=n\boldsymbol{\pi}+(-1)^n\dfrac{\boldsymbol{\pi}}{6}}}$