Answer
Verified
477.9k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE