Math, asked by divya3356, 1 year ago

2cosSquare X - 5 sin x + 1 is equal to zero solve it

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Answered by Anonymous
1
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〽Actually welcome to the concept of the Trigonometric functions ....

〽After solving we get .....
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Answered by Anonymous
29

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}}}

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