Solve the trigonometric expression $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)=2$.
Answer
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Hint: We will apply the formula given by $\sin \left( 2x \right)=2\sin x\cos x$. This is a trigonometric formula. We will use it to evaluate the trigonometric expression that is given in the question. We will also use the square root formula which is given by $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the roots of a quadratic equation that we will come across during solving the question.
Complete step-by-step answer:
We will first consider the trigonometric expression given to us as $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)=2...(i)$. We will take the 2 term to the left side of the equation (i) which is given by $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=0$. This is because we are here to find the value of x and in that case we will take all the terms to the left side of the equation. As we can clearly see that there is a double angle in the expression which is given as ${{\sin }^{2}}\left( 2x \right)$. So, our first step will be to use the formula $\sin \left( 2x \right)=2\sin x\cos x$ with the help of which we will evaluate the first step of the expression. Therefore, we get
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+{{\left( \sin \left( 2x \right) \right)}^{2}}-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+{{\left( 2\sin x\cos x \right)}^{2}}-2 \\
\end{align}$
After opening the square we get a new term which is $\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+{{\left( 2\sin x\cos x \right)}^{2}}-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x-2 \\
\end{align}$
Now we have the same angle x in both expressions. Therefore, we will take the square of sine term and 2 as common value. Thus, we will proceed as,
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2{{\cos }^{2}}x \right)-2 \\
\end{align}$
Now we will use the formula here which is given by ${{\cos }^{2}}x=1-{{\sin }^{2}}x$. Therefore, we get
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2{{\cos }^{2}}x \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2\left( 1-{{\sin }^{2}}x \right) \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2-2{{\sin }^{2}}x \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 3-2{{\sin }^{2}}x \right)-2 \\
\end{align}$
If we take ${{\sin }^{2}}x={{x}^{2}}$ in the right side of the equation then, we will get
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 3-2{{\sin }^{2}}x \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{x}^{2}}\left( 3-2{{x}^{2}} \right)-2 \\
\end{align}$
After multiplying the right side of the equation with each other we will get
$\begin{align}
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{x}^{2}}\left( 3-2{{x}^{2}} \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=6{{x}^{2}}-4{{x}^{4}}-2 \\
\end{align}$
As we have $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=0$ therefore, we can also write $6{{x}^{2}}-4{{x}^{4}}-2=0$. Now we will substitute ${{x}^{2}}=y$. Therefore our equation changes to $6y-4{{y}^{2}}-2=0$. Now we will use the formula of square root which is given by $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By substituting a = - 4, b = 6 and c = -2. Thus we have
$\begin{align}
& y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& \Rightarrow y=\dfrac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\left( -4 \right)\left( -2 \right)}}{2\left( -4 \right)} \\
& \Rightarrow y=\dfrac{-6\pm \sqrt{36-32}}{-8} \\
& \Rightarrow y=\dfrac{-6\pm \sqrt{4}}{-8} \\
& \Rightarrow y=\dfrac{-6\pm 2}{-8} \\
& \Rightarrow y=\dfrac{-3\pm 1}{-4} \\
\end{align}$
Thus we get $y=\dfrac{-3+1}{-4}$ and $y=\dfrac{-3-1}{-4}$. Which further gives $y=\dfrac{-2}{-4}$ and $y=\dfrac{-4}{-4}$. So, the values of y are $\dfrac{1}{2}$ and 1. Since ${{x}^{2}}=y$ and ${{\sin }^{2}}x={{x}^{2}}$ therefore we have ${{\sin }^{2}}x=\dfrac{1}{2}$ and ${{\sin }^{2}}x=1$. Thus, we have $\sin x=\dfrac{1}{\sqrt{2}}$ and $\sin x=1$.
Now we will consider $\sin x=\dfrac{1}{\sqrt{2}}$. We know that the value of $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$. As we know that the value of sine is positive in the first and the second quadrant. So, in the first quadrant we will have $\sin x=\sin \left( \dfrac{\pi }{4} \right)$. As in the first quadrant $x=\theta $ therefore we get $x=\dfrac{\pi }{4}$.
Also in the second quadrant we will have $\sin x=\sin \left( \dfrac{\pi }{4} \right)$. As in the second quadrant $x=\pi -\theta $ therefore we get
$\begin{align}
& x=\pi -\dfrac{\pi }{4} \\
& \Rightarrow x=\dfrac{4\pi -\pi }{4} \\
& \Rightarrow x=\dfrac{3\pi }{4} \\
\end{align}$
Now we will consider $\sin x=1$. We know that the value of $\sin \left( \dfrac{\pi }{2} \right)=1$. As we know that the value of sine is positive in the first and the second quadrant. So, in the first quadrant we will have $\sin x=\sin \left( \dfrac{\pi }{2} \right)$. As in the first quadrant $x=\theta $ therefore we get $x=\dfrac{\pi }{2}$.
Also in the second quadrant we will have $\sin x=1$. As in the second quadrant $x=\pi -\theta $ therefore we get
$\begin{align}
& x=\pi -\dfrac{\pi }{2} \\
& \Rightarrow x=\dfrac{2\pi -\pi }{2} \\
& \Rightarrow x=\dfrac{\pi }{2} \\
\end{align}$
Hence the values of x are $x=\dfrac{\pi }{4},x=\dfrac{3\pi }{4},x=\dfrac{\pi }{2}$ and $x=\dfrac{\pi }{2}$.
Note: We can also solve the quadratic equation $6y-4{{y}^{2}}-2=0$ with the help of hit and trial method. As with this method we can find the roots of y. This method works by substituting the integers one by one. After that we select those which satisfy the equation. In case we get one value satisfying the equation so we will then divide the equation by the factor in order to stop putting every number in the equation. As we know that ${{\sin }^{2}}x=\dfrac{1}{2}$ and ${{\sin }^{2}}x=1$. Thus, we actually got the values $\sin x=\pm \dfrac{1}{\sqrt{2}}$ and $\sin x=\pm 1$. But we have rejected the negative values and wrote $\sin x=\dfrac{1}{\sqrt{2}}$ and $\sin x=1$ directly.
Complete step-by-step answer:
We will first consider the trigonometric expression given to us as $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)=2...(i)$. We will take the 2 term to the left side of the equation (i) which is given by $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=0$. This is because we are here to find the value of x and in that case we will take all the terms to the left side of the equation. As we can clearly see that there is a double angle in the expression which is given as ${{\sin }^{2}}\left( 2x \right)$. So, our first step will be to use the formula $\sin \left( 2x \right)=2\sin x\cos x$ with the help of which we will evaluate the first step of the expression. Therefore, we get
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+{{\left( \sin \left( 2x \right) \right)}^{2}}-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+{{\left( 2\sin x\cos x \right)}^{2}}-2 \\
\end{align}$
After opening the square we get a new term which is $\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+{{\left( 2\sin x\cos x \right)}^{2}}-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x-2 \\
\end{align}$
Now we have the same angle x in both expressions. Therefore, we will take the square of sine term and 2 as common value. Thus, we will proceed as,
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x+4{{\sin }^{2}}x{{\cos }^{2}}x-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2{{\cos }^{2}}x \right)-2 \\
\end{align}$
Now we will use the formula here which is given by ${{\cos }^{2}}x=1-{{\sin }^{2}}x$. Therefore, we get
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2{{\cos }^{2}}x \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2\left( 1-{{\sin }^{2}}x \right) \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 1+2-2{{\sin }^{2}}x \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 3-2{{\sin }^{2}}x \right)-2 \\
\end{align}$
If we take ${{\sin }^{2}}x={{x}^{2}}$ in the right side of the equation then, we will get
$\begin{align}
& 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{\sin }^{2}}x\left( 3-2{{\sin }^{2}}x \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{x}^{2}}\left( 3-2{{x}^{2}} \right)-2 \\
\end{align}$
After multiplying the right side of the equation with each other we will get
$\begin{align}
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=2{{x}^{2}}\left( 3-2{{x}^{2}} \right)-2 \\
& \Rightarrow 2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=6{{x}^{2}}-4{{x}^{4}}-2 \\
\end{align}$
As we have $2{{\sin }^{2}}x+{{\sin }^{2}}\left( 2x \right)-2=0$ therefore, we can also write $6{{x}^{2}}-4{{x}^{4}}-2=0$. Now we will substitute ${{x}^{2}}=y$. Therefore our equation changes to $6y-4{{y}^{2}}-2=0$. Now we will use the formula of square root which is given by $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By substituting a = - 4, b = 6 and c = -2. Thus we have
$\begin{align}
& y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& \Rightarrow y=\dfrac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\left( -4 \right)\left( -2 \right)}}{2\left( -4 \right)} \\
& \Rightarrow y=\dfrac{-6\pm \sqrt{36-32}}{-8} \\
& \Rightarrow y=\dfrac{-6\pm \sqrt{4}}{-8} \\
& \Rightarrow y=\dfrac{-6\pm 2}{-8} \\
& \Rightarrow y=\dfrac{-3\pm 1}{-4} \\
\end{align}$
Thus we get $y=\dfrac{-3+1}{-4}$ and $y=\dfrac{-3-1}{-4}$. Which further gives $y=\dfrac{-2}{-4}$ and $y=\dfrac{-4}{-4}$. So, the values of y are $\dfrac{1}{2}$ and 1. Since ${{x}^{2}}=y$ and ${{\sin }^{2}}x={{x}^{2}}$ therefore we have ${{\sin }^{2}}x=\dfrac{1}{2}$ and ${{\sin }^{2}}x=1$. Thus, we have $\sin x=\dfrac{1}{\sqrt{2}}$ and $\sin x=1$.
Now we will consider $\sin x=\dfrac{1}{\sqrt{2}}$. We know that the value of $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$. As we know that the value of sine is positive in the first and the second quadrant. So, in the first quadrant we will have $\sin x=\sin \left( \dfrac{\pi }{4} \right)$. As in the first quadrant $x=\theta $ therefore we get $x=\dfrac{\pi }{4}$.
Also in the second quadrant we will have $\sin x=\sin \left( \dfrac{\pi }{4} \right)$. As in the second quadrant $x=\pi -\theta $ therefore we get
$\begin{align}
& x=\pi -\dfrac{\pi }{4} \\
& \Rightarrow x=\dfrac{4\pi -\pi }{4} \\
& \Rightarrow x=\dfrac{3\pi }{4} \\
\end{align}$
Now we will consider $\sin x=1$. We know that the value of $\sin \left( \dfrac{\pi }{2} \right)=1$. As we know that the value of sine is positive in the first and the second quadrant. So, in the first quadrant we will have $\sin x=\sin \left( \dfrac{\pi }{2} \right)$. As in the first quadrant $x=\theta $ therefore we get $x=\dfrac{\pi }{2}$.
Also in the second quadrant we will have $\sin x=1$. As in the second quadrant $x=\pi -\theta $ therefore we get
$\begin{align}
& x=\pi -\dfrac{\pi }{2} \\
& \Rightarrow x=\dfrac{2\pi -\pi }{2} \\
& \Rightarrow x=\dfrac{\pi }{2} \\
\end{align}$
Hence the values of x are $x=\dfrac{\pi }{4},x=\dfrac{3\pi }{4},x=\dfrac{\pi }{2}$ and $x=\dfrac{\pi }{2}$.
Note: We can also solve the quadratic equation $6y-4{{y}^{2}}-2=0$ with the help of hit and trial method. As with this method we can find the roots of y. This method works by substituting the integers one by one. After that we select those which satisfy the equation. In case we get one value satisfying the equation so we will then divide the equation by the factor in order to stop putting every number in the equation. As we know that ${{\sin }^{2}}x=\dfrac{1}{2}$ and ${{\sin }^{2}}x=1$. Thus, we actually got the values $\sin x=\pm \dfrac{1}{\sqrt{2}}$ and $\sin x=\pm 1$. But we have rejected the negative values and wrote $\sin x=\dfrac{1}{\sqrt{2}}$ and $\sin x=1$ directly.
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