
$2{{\sin }^{2}}x+{{\sin }^{2}}2x=2$, $-\pi<{x}<\pi$ then $x=$.
A. $\pm \dfrac{\pi }{6}$
B. $\pm \dfrac{\pi }{4}$
C. $\dfrac{3\pi }{2}$
D. None of these.
Answer
162.6k+ views
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 $-\piWe 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 $-\piThe 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.
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
$2{{\sin }^{2}}x+{{\sin }^{2}}2x=2$
$2{{\sin }^{2}}x+{{\left( 2\sin x\cos x \right)}^{2}}=2$
$2{{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x=2$
${{\sin }^{2}}x+2{{\sin }^{2}}x{{\cos }^{2}}x=1$
$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
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.
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