Question

If \[5\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0\], $-\pi <\theta <\pi $, then $\theta =$
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{3},{{\cos }^{-1}}\dfrac{3}{5}\]
C. \[{{\cos }^{-1}}\dfrac{3}{5}\]
D. \[\dfrac{\pi }{3},\pi -{{\cos }^{-1}}\dfrac{3}{5}\]

Answer 1

Hint: To find the value of $\theta $, we will simplify the equation using the double angle formula of cos. After this we will rewrite the equation in a way that we can form factors for the equation. Then we will equate both the factors to zero and find the value $\theta $ using trigonometric table of values of function at different angles.

Formula Used: $\cos 2\theta =2{{\cos }^{2}}\theta -1$

Complete step by step solution: We are given a trigonometric equation \[5\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0\] where $-\pi <\theta <\pi $ and we have to derive the value of $\theta $.
We will use the double angle formula of cos that is $\cos 2A$and then simplify the equation.
\[\begin{align}
  & 5\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0 \\
 & 5\left( 2{{\cos }^{2}}\theta -1 \right)+2{{\cos }^{2}}\dfrac{\theta }{2}+1=0 \\
 & 10{{\cos }^{2}}\theta -4+2{{\cos }^{2}}\dfrac{\theta }{2}=0
\end{align}\]
We will again apply the double angle formula of cos.
\[\begin{align}
  & 10{{\cos }^{2}}\theta -3+\left( 2{{\cos }^{2}}\dfrac{\theta }{2}-1 \right)=0 \\
 & 10{{\cos }^{2}}\theta +\cos \theta -3=0
\end{align}\]
We will now write $\cos \theta $ as $\cos \theta =6\cos \theta -5\cos \theta $ to form the factors.
\[\begin{align}
  & 10{{\cos }^{2}}\theta +6\cos \theta -5\cos \theta -3=0 \\
 & 10{{\cos }^{2}}\theta -5\cos \theta +6\cos \theta -3=0 \\
 & 5\cos \theta (2\cos \theta -1)+3(2\cos \theta -1)=0 \\
 & (2\cos \theta -1)(5\cos \theta +3)=0
\end{align}\]
Now we will equate both the factors to zero.
\[\begin{align}
  & 2\cos \theta -1=0 \\
 & \cos \theta =\dfrac{1}{2}
\end{align}\] or \[\begin{align}
  & 5\cos \theta +3=0 \\
 & \cos \theta =\dfrac{-3}{5}
\end{align}\]
We know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$. So,
\[\begin{align}
  & \cos \theta =\cos \dfrac{\pi }{3} \\
 & \theta =\dfrac{\pi }{3}
\end{align}\] or $\cos $\[\begin{align}
  & \cos \theta =\dfrac{-3}{5} \\
 & \theta ={{\cos }^{-1}}\left( \dfrac{-3}{5} \right) \\
\end{align}\]
\[\theta ={{\cos }^{-1}}\left( \dfrac{-3}{5} \right)\] will reside in the second quadrant because we are given that value of $\theta $ lies in the interval $-\pi <\theta <\pi $ so the value will be \[\theta =\pi -{{\cos }^{-1}}\left( \dfrac{3}{5} \right)\].
The value of $\theta $ for the trigonometric equation \[5\cos 2\theta +2{{\cos }^{2}}\dfrac{\theta }{2}+1=0\] when $-\pi <\theta <\pi $ is \[\theta =\dfrac{\pi }{3},\pi -{{\cos }^{-1}}\left( \dfrac{3}{5} \right)\].

Option ‘D’ is correct

Note: There are four quadrants for trigonometric functions in which they have specific values and signs. In the first quadrant the angle is less than or equal to ${{90}^{0}}$ represented as $\dfrac{\pi }{2}-\theta $ and all the functions are positive.
In the second quadrant only function sin and cosec is positive and the angle is greater than ${{90}^{0}}$ but less than ${{180}^{0}}$represented as $\dfrac{\pi }{2}+\theta ,\pi -\theta $.
In the third quadrant, only function tan and cot is positive and the angle is greater than ${{180}^{0}}$ but less than ${{270}^{0}}$represented as $\pi +\theta ,\dfrac{3\pi }{2}-\theta $.
In the fourth quadrant, only function cos and sec is positive and the angle is greater than ${{270}^{0}}$ but less than ${{360}^{0}}$represented as $\dfrac{3\pi }{2}+\theta ,2\pi -\theta $.