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$2{{\sin }^{2}}\theta =3\cos \theta $ in the interval $0\le \theta \le 2\pi $.

Answer
Verified

Hint: Use the formula \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\] in the given equation and convert the equation into quadratic equation in the form of \[\cos \theta \]. Find the solution of the quadratic equation and use the concept “If \[\cos x=\cos y\] then, \[x=2n\pi \pm y\]” so that you can get the answer in the interval of $0\le \theta \le 2\pi $.

Complete step-by-step answer:

To solve the above equation we have to write it down therefore,

$2{{\sin }^{2}}\theta =3\cos \theta $

As we can see the ${{\sin }^{2}}\theta $ can easily be converted to ${{\cos }^{2}}\theta $ by using the formula given below so that we will find easier to solve the equation in \[\cos \theta \].

Formula:

\[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]

If we use the above formula in the given equation w will get,

\[\therefore 2\left( 1-{{\cos }^{2}}\theta \right)=3\cos \theta \]

If we multiply by ‘2’ inside the bracket we will get,

\[\therefore 2-2{{\cos }^{2}}\theta =3\cos \theta \]

If we shift \[3\cos \theta \] on the left hand side of the equation we will get,

\[\therefore 2-2{{\cos }^{2}}\theta -3\cos \theta =0\]

Now we will multiply the above equation by ‘-1’ to get,

\[\therefore \left( -1 \right)\times \left( 2-2{{\cos }^{2}}\theta -3\cos \theta \right)=0\]

\[\therefore -2+2{{\cos }^{2}}\theta +3\cos \theta =0\]

By rearranging the above equation we will get,

\[\therefore 2{{\cos }^{2}}\theta +3\cos \theta -2=0\]

Now we can write \[3\cos \theta \]as \[\left( -4\cos \theta +\cos \theta \right)\] therefore we will get,

\[\therefore 2{{\cos }^{2}}\theta +\left( -4\cos \theta +\cos \theta \right)-2=0\]

\[\therefore 2{{\cos }^{2}}\theta -4\cos \theta +\cos \theta -2=0\]

Taking \[2\cos \theta \] common from above equation we will get,

\[\therefore 2\cos \theta \left( \cos \theta -2 \right)+\cos \theta -2=0\]

If we rearrange the above equation we will get,

\[\therefore 2\cos \theta \left( \cos \theta -2 \right)+\left( \cos \theta -2 \right)=0\]

Also by taking \[\left( \cos \theta -2 \right)\] from the above equation we will get,

\[\therefore \left( \cos \theta -2 \right)\left( 2\cos \theta +1 \right)=0\]

As they are in multiplication therefore we can write,

\[\left( \cos \theta -2 \right)=0\] ………………………….. (1)

OR

\[\left( 2\cos \theta +1 \right)=0\]………………………….. (2)

We will write the equation (1) first, therefore,

\[\left( \cos \theta -2 \right)=0\]

\[\therefore \cos \theta =2\]

But as we know that the range of \[\cos \theta \] is [-1, 1] Therefore the value of \[\cos \theta \] equal to 2 is not possible, therefore we can write,

\[\therefore \cos \theta =2\] Not possible.

Hence rejected.

Now we will solve the equation (2), therefore,

\[\left( 2\cos \theta +1 \right)=0\]

\[\therefore 2\cos \theta =-1\]

\[\therefore \cos \theta =\dfrac{-1}{2}\]

\[\therefore \cos \theta =-\dfrac{1}{2}\]

As we know that the value of \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] and if we put this value in above equation we will get,

\[\therefore \cos \theta =-\cos \dfrac{\pi }{3}\]

Now to proceed further in the solution we should know the formula given below,

Formula:

\[-\cos x=\cos \left( \pi -x \right)\]

If we use the formula in the equation we will get,

\[\therefore \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right)\]

\[\therefore \cos \theta =\cos \left( \dfrac{3\pi -\pi }{3} \right)\]

\[\therefore \cos \theta =\cos \left( \dfrac{2\pi }{3} \right)\]

To proceed further in the solution we should know the formula given below,

Formula:

If \[\cos x=\cos y\] then, \[x=2n\pi \pm y\]

If we use the above formula in the equation we will get,

As, \[\cos \theta =\cos \left( \dfrac{2\pi }{3} \right)\] therefore, \[\theta =2n\pi \pm \dfrac{2\pi }{3}\]

As we have to find the solution between \[0\le \theta \le 2\pi \] therefore we will put n = 1 in the solution, therefore we will get,

\[\therefore \theta =2\left( 1 \right)\pi \pm \dfrac{2\pi }{3}\]

This can also be written as,

\[\therefore \theta =2\pi +\dfrac{2\pi }{3}\] And \[\therefore \theta =2\pi -\dfrac{2\pi }{3}\]

If we do the required calculations we will get,

\[\therefore \theta =\dfrac{6\pi +2\pi }{3}\] And \[\therefore \theta =\dfrac{6\pi -2\pi }{3}\]

\[\therefore \theta =\dfrac{8\pi }{3}\] And \[\therefore \theta =\dfrac{4\pi }{3}\]

As we know that \[\dfrac{8\pi }{3}>2\pi \] therefore the answer is,

\[\therefore \theta =\dfrac{4\pi }{3}\]

Therefore the final answer is,

\[\therefore \theta =\dfrac{2\pi }{3},\dfrac{4\pi }{3}\]

Therefore the solution of $2{{\sin }^{2}}\theta =3\cos \theta $ in the interval $0\le \theta \le 2\pi $ is \[\dfrac{2\pi }{3}\] and \[\dfrac{4\pi }{3}\].

Note: From the step \[\cos \theta =-\dfrac{1}{2}\] you can also solve further by using the formula of inverse trigonometric functions i.e. \[{{\cos }^{-1}}\left( -x \right)=\pi ={{\cos }^{-1}}x\] and you can further manage the solution in the given interval.

Complete step-by-step answer:

To solve the above equation we have to write it down therefore,

$2{{\sin }^{2}}\theta =3\cos \theta $

As we can see the ${{\sin }^{2}}\theta $ can easily be converted to ${{\cos }^{2}}\theta $ by using the formula given below so that we will find easier to solve the equation in \[\cos \theta \].

Formula:

\[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]

If we use the above formula in the given equation w will get,

\[\therefore 2\left( 1-{{\cos }^{2}}\theta \right)=3\cos \theta \]

If we multiply by ‘2’ inside the bracket we will get,

\[\therefore 2-2{{\cos }^{2}}\theta =3\cos \theta \]

If we shift \[3\cos \theta \] on the left hand side of the equation we will get,

\[\therefore 2-2{{\cos }^{2}}\theta -3\cos \theta =0\]

Now we will multiply the above equation by ‘-1’ to get,

\[\therefore \left( -1 \right)\times \left( 2-2{{\cos }^{2}}\theta -3\cos \theta \right)=0\]

\[\therefore -2+2{{\cos }^{2}}\theta +3\cos \theta =0\]

By rearranging the above equation we will get,

\[\therefore 2{{\cos }^{2}}\theta +3\cos \theta -2=0\]

Now we can write \[3\cos \theta \]as \[\left( -4\cos \theta +\cos \theta \right)\] therefore we will get,

\[\therefore 2{{\cos }^{2}}\theta +\left( -4\cos \theta +\cos \theta \right)-2=0\]

\[\therefore 2{{\cos }^{2}}\theta -4\cos \theta +\cos \theta -2=0\]

Taking \[2\cos \theta \] common from above equation we will get,

\[\therefore 2\cos \theta \left( \cos \theta -2 \right)+\cos \theta -2=0\]

If we rearrange the above equation we will get,

\[\therefore 2\cos \theta \left( \cos \theta -2 \right)+\left( \cos \theta -2 \right)=0\]

Also by taking \[\left( \cos \theta -2 \right)\] from the above equation we will get,

\[\therefore \left( \cos \theta -2 \right)\left( 2\cos \theta +1 \right)=0\]

As they are in multiplication therefore we can write,

\[\left( \cos \theta -2 \right)=0\] ………………………….. (1)

OR

\[\left( 2\cos \theta +1 \right)=0\]………………………….. (2)

We will write the equation (1) first, therefore,

\[\left( \cos \theta -2 \right)=0\]

\[\therefore \cos \theta =2\]

But as we know that the range of \[\cos \theta \] is [-1, 1] Therefore the value of \[\cos \theta \] equal to 2 is not possible, therefore we can write,

\[\therefore \cos \theta =2\] Not possible.

Hence rejected.

Now we will solve the equation (2), therefore,

\[\left( 2\cos \theta +1 \right)=0\]

\[\therefore 2\cos \theta =-1\]

\[\therefore \cos \theta =\dfrac{-1}{2}\]

\[\therefore \cos \theta =-\dfrac{1}{2}\]

As we know that the value of \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] and if we put this value in above equation we will get,

\[\therefore \cos \theta =-\cos \dfrac{\pi }{3}\]

Now to proceed further in the solution we should know the formula given below,

Formula:

\[-\cos x=\cos \left( \pi -x \right)\]

If we use the formula in the equation we will get,

\[\therefore \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right)\]

\[\therefore \cos \theta =\cos \left( \dfrac{3\pi -\pi }{3} \right)\]

\[\therefore \cos \theta =\cos \left( \dfrac{2\pi }{3} \right)\]

To proceed further in the solution we should know the formula given below,

Formula:

If \[\cos x=\cos y\] then, \[x=2n\pi \pm y\]

If we use the above formula in the equation we will get,

As, \[\cos \theta =\cos \left( \dfrac{2\pi }{3} \right)\] therefore, \[\theta =2n\pi \pm \dfrac{2\pi }{3}\]

As we have to find the solution between \[0\le \theta \le 2\pi \] therefore we will put n = 1 in the solution, therefore we will get,

\[\therefore \theta =2\left( 1 \right)\pi \pm \dfrac{2\pi }{3}\]

This can also be written as,

\[\therefore \theta =2\pi +\dfrac{2\pi }{3}\] And \[\therefore \theta =2\pi -\dfrac{2\pi }{3}\]

If we do the required calculations we will get,

\[\therefore \theta =\dfrac{6\pi +2\pi }{3}\] And \[\therefore \theta =\dfrac{6\pi -2\pi }{3}\]

\[\therefore \theta =\dfrac{8\pi }{3}\] And \[\therefore \theta =\dfrac{4\pi }{3}\]

As we know that \[\dfrac{8\pi }{3}>2\pi \] therefore the answer is,

\[\therefore \theta =\dfrac{4\pi }{3}\]

Therefore the final answer is,

\[\therefore \theta =\dfrac{2\pi }{3},\dfrac{4\pi }{3}\]

Therefore the solution of $2{{\sin }^{2}}\theta =3\cos \theta $ in the interval $0\le \theta \le 2\pi $ is \[\dfrac{2\pi }{3}\] and \[\dfrac{4\pi }{3}\].

Note: From the step \[\cos \theta =-\dfrac{1}{2}\] you can also solve further by using the formula of inverse trigonometric functions i.e. \[{{\cos }^{-1}}\left( -x \right)=\pi ={{\cos }^{-1}}x\] and you can further manage the solution in the given interval.

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