Question

$2{{\sin }^{2}}x+{{\sin }^{2}}2x=2$, $-\pi<{x}<\pi$ then $x=$.
A. $\pm \dfrac{\pi }{6}$
B. $\pm \dfrac{\pi }{4}$
C. $\dfrac{3\pi }{2}$
D. None of these.

Answer 1

Hint: To find the value of $x$ we will simplify the given equation using the trigonometric formulas of $\sin 2A$ and ${{\cos }^{2}}x$. After simplifying the equation, we will get two factors where we will apply the theorem of general equations for sin and cos.
The general equation of angle when $\sin x=\sin y$ for any real number $x$ and $y$is $x=n\pi +{{(-1)}^{n}}y$ and the general equation when $\cos x=0$ for any of the real number $x$ is $x=(2n+1)\dfrac{\pi }{2}$where $n\in Z$.

Formula Used:$\sin 2A=2\sin A\cos A$

Complete step by step solution: We are given a trigonometric equation $2{{\sin }^{2}}x+{{\sin }^{2}}2x=2$ where $-\pi We will take the given equation and apply the formula of $\sin 2A$.
$2{{\sin }^{2}}x+{{\sin }^{2}}2x=2$
$2{{\sin }^{2}}x+{{\left( 2\sin x\cos x \right)}^{2}}=2$
$2{{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x=2$
${{\sin }^{2}}x+2{{\sin }^{2}}x{{\cos }^{2}}x=1$
$2{{\sin }^{2}}x{{\cos }^{2}}x+{{\sin }^{2}}x-1=0 $


We will now simplify the equation so that we can apply the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ in the equation.
  $2{{\sin }^{2}}x{{\cos }^{2}}x-(1-{{\sin }^{2}}x)=0 $
 $2{{\sin }^{2}}x{{\cos }^{2}}x-{{\cos }^{2}}x=0 $
  ${{\cos }^{2}}x(2{{\sin }^{2}}x-1)=0 $
We will now equate both the factors to zero.
${{\cos }^{2}}x=0$ or
  $2{{\sin }^{2}}x-1=0$
  ${{\sin }^{2}}x=\dfrac{1}{2} $
We know that ${{\sin }^{2}}\dfrac{\pi }{4}=\dfrac{1}{2}$. So,
 ${{\cos }^{2}}x=0$
 $cos x=0 $
 or
  ${{\sin }^{2}}x={{\sin }^{2}}\dfrac{\pi }{4} $
  $\sin x=\sin \dfrac{\pi }{4} $
We will now apply both the theorems of general equation here,
$x=(2n+1)\dfrac{\pi }{2}$ or $x=n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}$
We have derived the general solutions here and to derive the principal solutions we will substitute the integer values of $n$.
Substituting the value of $n=-2,-1,0,1,2$.
$x=\dfrac{-3\pi }{2},\dfrac{-\pi }{2},\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2}$ or $x=\dfrac{-7\pi }{4},\dfrac{-5\pi }{4},\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{9\pi }{4}$
As we are given that value of $x$ will lie in the interval of $-\pi The possible value of $x$ in the interval of $-\pi The value of $x$ for the trigonometric equation $2{{\sin }^{2}}x+{{\sin }^{2}}2x=2$ when $-\pi<{x}<\pi$ is $x=\pm \dfrac{\pi }{4}$. Hence the correct option is (B).

Option ‘B’ is correct

Note: We could have simplified the given equation by using any other trigonometric formulas. But after equating the factors even if the factors are different we will get the similar result.