Question

Find the number of integral values of n so that \[\sin x\left( \sin x+\cos x \right)=n\] has at least one solution.

Answer 1

Hint: Now we are given with function \[\sin x\left( \sin x+\cos x \right)=n\] . We will open the brackets and then use the formula ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ and $2\sin x\cos x=\sin 2x$ to rewrite the equation accordingly. Now we will rearrange the terms so that we get the equation in the form of $\sin 2x$ and $\cos 2x$ . Now we will divide the whole equation by $\sqrt{2}$ . Now we know that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ Hence we will substitute this values in the equation obtained. Now in the new equation formed we will use the formula $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ to simplify the equation. Finally we will use the fact that for any angle A. $-1\le \sin A\le 1$ . Hence we will get the inequality in n which we will solve to find possible integral values of n.

Complete step-by-step solution:
Now first consider the given function \[\sin x\left( \sin x+\cos x \right)=n\] .
Opening the bracket in the above equation we get ${{\sin }^{2}}x+\sin x\cos x=n$
Now we know that ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ and $2\sin x\cos x=\sin 2x$ .
This means ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ and $\sin x\cos x=\dfrac{\sin 2x}{2}$
Now substituting this we get $\dfrac{1-\cos 2x}{2}+\dfrac{\sin 2x}{2}=n$
Hence we have $\dfrac{1-\cos 2x+\sin 2x}{2}=n$
Now rearranging the given terms we get $\sin 2x-\cos 2x=2n-1$ .
Let us divide the whole equation by $\sqrt{2}$ Hence we get
$\dfrac{1}{\sqrt{2}}\sin 2x-\dfrac{1}{\sqrt{2}}\cos 2x=\dfrac{2n-1}{\sqrt{2}}$
Now we have that the values of $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$
Hence substituting this in the above equation we get.
$\cos \left( \dfrac{\pi }{4} \right)\sin 2x-\sin \left( \dfrac{\pi }{4} \right)\cos 2x=\dfrac{2n-1}{\sqrt{2}}$
Now we know that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ . Hence using this formula we will simplify the equation to get.
\[\sin \left( 2x-\dfrac{\pi }{4} \right)=\dfrac{2n-1}{\sqrt{2}}.......................\left( 1 \right)\]
Now we know that $-1\le \sin \left( 2x-\dfrac{\pi }{4} \right)\le 1$
Hence from equation (1) we get.
\[-1\le \dfrac{2n-1}{\sqrt{2}}<1\]
\[\begin{align}
  & \Rightarrow -\sqrt{2}\le 2n-1\le \sqrt{2} \\
 & \Rightarrow 1-\sqrt{2}\le 2n\le \sqrt{2}+1 \\
 & \Rightarrow \dfrac{1-\sqrt{2}}{2}\le n\le \dfrac{\sqrt{2}+1}{2} \\
\end{align}\]
Now we have $\sqrt{2}=1.41$ Now substituting this we get.
$\begin{align}
  & \dfrac{1-1.41}{2}\le n\le \dfrac{1+1.41}{2} \\
 & \Rightarrow \dfrac{-0.41}{2}\le n\le \dfrac{2.41}{2} \\
 & \Rightarrow -0.205\le n\le 1.205 \\
\end{align}$
Hence the only possible integral values of n is 0 and 1. Hence only 2 Integral values of n are possible.

Note: Now instead of using the range of sin to form inequality we can also find the maximum and minimum value of $\sin 2x-\cos 2x$ by differentiating and equating it to zero. Hence we will get inequality in n again and we can solve it to find the required integral values.