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

The equation sin⁡x(sin⁡x+cos⁡x)=k\sin x\left( \sin x+\cos x \right)=k has real solutions, where ‘k’ is a real number. Then
(a) 0≤k≤1+220\le k\le \dfrac{1+\sqrt{2}}{2}
(b) 2−3≤k≤2+32-\sqrt{3}\le k\le 2+\sqrt{3}
(c) 0≤k≤2−30\le k\le 2-\sqrt{3}
(d) 1−22≤k≤1+22\dfrac{1-\sqrt{2}}{2}\le k\le \dfrac{1+\sqrt{2}}{2}

Explanation

Solution

Hint: Try to use the following identities such as cos⁡2x=1−2sin⁡2x,sin⁡2x=2sin⁡xcos⁡x\cos 2x=1-2{{\sin }^{2}}x,{{\sin }^{2}}x=2\sin x\cos x and lastly, −a2+b2≤asin⁡θ+bcos⁡θ≤a2+b2-\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(sin⁡x+cos⁡x)=k...........(i)\sin x\left( \sin x+\cos x \right)=k...........\left( i \right)
Now, we will expand the equation (i), we will get;
k=sin⁡2x+sin⁡xcos⁡x..............(ii)k={{\sin }^{2}}x+\sin x\cos x..............\left( ii \right)
Now, we know the identity,
cos⁡2x=1−2sin⁡2x\cos 2x=1-2{{\sin }^{2}}x
Which can also be written as
2sin⁡2x=1−cos⁡2x2{{\sin }^{2}}x=1-\cos 2x
So, sin⁡2x=1−cos⁡2x2...........(iii){{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}...........\left( iii \right)
Now, we will use another identity;
sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x
Which can also be formed as;
sin⁡xcos⁡x=sin⁡2x2...........(iv)\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;
k=12−cos⁡2x2+sin⁡2x2 ⇒k=12+sin⁡2x2−cos⁡2x2.............(v) \begin{aligned} & 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{aligned}
Now, we will use another identity which is;
−a2+b2≤asin⁡θ+bcos⁡θ≤a2+b2-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin \theta +b\cos \theta \le \sqrt{{{a}^{2}}+{{b}^{2}}}
We can replace θ\theta by 2x2x and 12\dfrac{1}{2} in place of aa and −12-\dfrac{1}{2} in place of bb.
We get;
⇒−(12)2+(−12)2≤(12)sin⁡(2x)+(−12)cos⁡2x≤(12)2+(−12)2\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

& \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{aligned}$$ 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.