# The equation $\sin x\left( \sin x+\cos x \right)=k$ has real solutions, where ‘k’ is a real number. Then

(a) $0\le k\le \dfrac{1+\sqrt{2}}{2}$

(b) $2-\sqrt{3}\le k\le 2+\sqrt{3}$

(c) $0\le k\le 2-\sqrt{3}$

(d) $\dfrac{1-\sqrt{2}}{2}\le k\le \dfrac{1+\sqrt{2}}{2}$

Answer

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Hint: Try to use the following identities such as $\cos 2x=1-2{{\sin }^{2}}x,{{\sin }^{2}}x=2\sin x\cos x$ and lastly, $-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin \theta +b\cos \theta \le \sqrt{{{a}^{2}}+{{b}^{2}}}$ to get the desired results.

Complete step-by-step answer:

In the question given, the equation is

$\sin x\left( \sin x+\cos x \right)=k...........\left( i \right)$

Now, we will expand the equation (i), we will get;

$k={{\sin }^{2}}x+\sin x\cos x..............\left( ii \right)$

Now, we know the identity,

$\cos 2x=1-2{{\sin }^{2}}x$

Which can also be written as

$2{{\sin }^{2}}x=1-\cos 2x$

So, ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}...........\left( iii \right)$

Now, we will use another identity;

$\sin 2x=2\sin x\cos x$

Which can also be formed as;

$\sin x\cos x=\dfrac{\sin 2x}{2}...........\left( iv \right)$

Now, substituting the results of equation (iii) and (iv) in equation (ii) we get;

$\begin{align}

& k=\dfrac{1}{2}-\dfrac{\cos 2x}{2}+\dfrac{\sin 2x}{2} \\

& \Rightarrow k=\dfrac{1}{2}+\dfrac{\sin 2x}{2}-\dfrac{\cos 2x}{2}.............\left( v \right) \\

\end{align}$

Now, we will use another identity which is;

$-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin \theta +b\cos \theta \le \sqrt{{{a}^{2}}+{{b}^{2}}}$

We can replace $\theta $ by $2x$ and $\dfrac{1}{2}$ in place of $a$ and $-\dfrac{1}{2}$ in place of $b$.

We get;

\[\Rightarrow -\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( -\dfrac{1}{2} \right)}^{2}}}\le \left( \dfrac{1}{2} \right)\sin \left( 2x \right)+\left( -\dfrac{1}{2} \right)\cos 2x\le \sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( -\dfrac{1}{2} \right)}^{2}}}\]

Solving this, we get

\[\begin{align}

& \Rightarrow -\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}\le \dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x\le \sqrt{\dfrac{1}{4}+\dfrac{1}{4}} \\

& \Rightarrow -\dfrac{1}{\sqrt{2}}\le \dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x\le \dfrac{1}{\sqrt{2}} \\

\end{align}\]

Now we will add $\dfrac{1}{2}$ to all the sides of inequality. We get;

$\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\le \dfrac{1}{2}+\dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x\le \dfrac{1}{2}+\dfrac{1}{\sqrt{2}}$

Now, we substitute the whole value of $\left( \dfrac{1}{2}+\dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x \right)$ by $'k'$ from equation (v), we get

$\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\le k\le \dfrac{1}{2}+\dfrac{1}{\sqrt{2}}$

Taking the LCM, we get

$\dfrac{1-\sqrt{2}}{2}\le k\le \dfrac{1+\sqrt{2}}{2}$

So, the answer is option (d).

Note: In these type of questions student generally get confused while converting $\sin x\cos x$ and ${{\sin }^{2}}x$ in the terms of $\sin 2x\text{ and }\cos 2x$ respectively.

Students generally don’t expand the given equation and start substituting values of different identities as it is, that is directly in the equation $\sin x\left( \sin x+\cos x \right)=k$. In this way they will get confused and will get the wrong answer.

Complete step-by-step answer:

In the question given, the equation is

$\sin x\left( \sin x+\cos x \right)=k...........\left( i \right)$

Now, we will expand the equation (i), we will get;

$k={{\sin }^{2}}x+\sin x\cos x..............\left( ii \right)$

Now, we know the identity,

$\cos 2x=1-2{{\sin }^{2}}x$

Which can also be written as

$2{{\sin }^{2}}x=1-\cos 2x$

So, ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}...........\left( iii \right)$

Now, we will use another identity;

$\sin 2x=2\sin x\cos x$

Which can also be formed as;

$\sin x\cos x=\dfrac{\sin 2x}{2}...........\left( iv \right)$

Now, substituting the results of equation (iii) and (iv) in equation (ii) we get;

$\begin{align}

& k=\dfrac{1}{2}-\dfrac{\cos 2x}{2}+\dfrac{\sin 2x}{2} \\

& \Rightarrow k=\dfrac{1}{2}+\dfrac{\sin 2x}{2}-\dfrac{\cos 2x}{2}.............\left( v \right) \\

\end{align}$

Now, we will use another identity which is;

$-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin \theta +b\cos \theta \le \sqrt{{{a}^{2}}+{{b}^{2}}}$

We can replace $\theta $ by $2x$ and $\dfrac{1}{2}$ in place of $a$ and $-\dfrac{1}{2}$ in place of $b$.

We get;

\[\Rightarrow -\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( -\dfrac{1}{2} \right)}^{2}}}\le \left( \dfrac{1}{2} \right)\sin \left( 2x \right)+\left( -\dfrac{1}{2} \right)\cos 2x\le \sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( -\dfrac{1}{2} \right)}^{2}}}\]

Solving this, we get

\[\begin{align}

& \Rightarrow -\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}\le \dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x\le \sqrt{\dfrac{1}{4}+\dfrac{1}{4}} \\

& \Rightarrow -\dfrac{1}{\sqrt{2}}\le \dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x\le \dfrac{1}{\sqrt{2}} \\

\end{align}\]

Now we will add $\dfrac{1}{2}$ to all the sides of inequality. We get;

$\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\le \dfrac{1}{2}+\dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x\le \dfrac{1}{2}+\dfrac{1}{\sqrt{2}}$

Now, we substitute the whole value of $\left( \dfrac{1}{2}+\dfrac{1}{2}\sin 2x-\dfrac{1}{2}\cos 2x \right)$ by $'k'$ from equation (v), we get

$\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\le k\le \dfrac{1}{2}+\dfrac{1}{\sqrt{2}}$

Taking the LCM, we get

$\dfrac{1-\sqrt{2}}{2}\le k\le \dfrac{1+\sqrt{2}}{2}$

So, the answer is option (d).

Note: In these type of questions student generally get confused while converting $\sin x\cos x$ and ${{\sin }^{2}}x$ in the terms of $\sin 2x\text{ and }\cos 2x$ respectively.

Students generally don’t expand the given equation and start substituting values of different identities as it is, that is directly in the equation $\sin x\left( \sin x+\cos x \right)=k$. In this way they will get confused and will get the wrong answer.

Last updated date: 19th Sep 2023

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