Question

The value of \[x\] between $0$ and $2\pi $ which satisfy the equation $\sin x\sqrt {8{{\cos }^2}x} = 1$ are in AP. Find the common difference of AP.

Answer 1

Hint: We are given with the equation $\sin x\sqrt {8{{\cos }^2}x} = 1$. Find the values for \[x\] in the equation $\sin x\sqrt {8{{\cos }^2}x} = 1$. We will get two conditions during solving for \[x\] which are $\cos x > 0$ and $\cos x < 0$ using these conditions of $\cos x$ find the values of $x$ between angles $0$ and $2\pi $ and then calculate the common difference between the values of $x$.

Complete Step by Step Solution:
From the question, we know that the values of $x$ must lie between $0$ and $2\pi $, so that they satisfy the equation $\sin x\sqrt {8{{\cos }^2}x} = 1$. Now, to find the values of $x$, we have to solve the equation $\sin x\sqrt {8{{\cos }^2}x} = 1$, so, we get –
$ \Rightarrow \sin x\sqrt {8{{\cos }^2}x} = 1$
We know that, $\sqrt 8 $ is equal to \[2\sqrt 2 \] and the square root of ${\cos ^2}x$ will be $\cos x$ of positive and negative both. Therefore, using these in the above equation, we get –
$ \Rightarrow 2\sqrt 2 \sin x\left| {\cos x} \right| = 1$
Using the transposition method and shifting $\sqrt 2 $ on right – hand side of the above equation, we get –
$ \Rightarrow 2\sin x\left| {\cos x} \right| = \dfrac{1}{{\sqrt 2 }} \cdots \left( 1 \right)$
As we know, the value of $\cos x$ can be positive and negative both.
Therefore, when $\cos x > 0$ or when it is positive, we can write the equation (1) as –
$ \Rightarrow 2\sin x\cos x = \dfrac{1}{{\sqrt 2 }}$
We know the identity of double angle formula of sine, $\sin 2x = 2\sin x\cos x$ , using this identity in the above equation, we get –
$
   \Rightarrow \sin 2x = \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow 2x = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
 $
From the question, we know that values of $x$ lie between 0 and $2\pi $. The value of $\sin $ is $\dfrac{1}{{\sqrt 2 }}$ when the angles are $\dfrac{\pi }{4},\dfrac{{3\pi }}{4}$ -
$ \Rightarrow 2x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4} \\
   \Rightarrow x = \dfrac{\pi }{8},\dfrac{{3\pi }}{8} \\ $
When $\cos x < 0$ or when it is negative then, the equation (1) can be written as –
$ \Rightarrow 2\sin x\cos x = - \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow \sin 2x = - \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow 2x = {\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) \\
   \Rightarrow 2x = \dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4} \\
   \Rightarrow x = \dfrac{{5\pi }}{8},\dfrac{{7\pi }}{8} \\
 $
Hence, now the required values of $x$ are $\dfrac{\pi }{8},\dfrac{{3\pi }}{8},\dfrac{{5\pi }}{8},\dfrac{{7\pi }}{8}$
According to the question, it is given that the values of $x$ are in AP so, to find the common difference of the AP, we have –
$
   \Rightarrow \dfrac{{3\pi }}{8} - \dfrac{\pi }{8} = \dfrac{{2\pi }}{8} \\
   \Rightarrow \dfrac{\pi }{4} \\
 $
Hence, the common difference of the AP is $\dfrac{\pi }{4}$.

Note:
When any series is in AP, then, if the second term is subtracted from first term, third term is subtracted from second term and it goes like this then we get the same difference in the AP. Trigonometric identities should be remembered by the student to solve any question having trigonometric terms.