Question

The expression $\cos 3\theta +\sin 3\theta +\left( 2\sin 2\theta -3 \right)\left( \sin \theta -\cos \theta \right)$ is positive for all $\theta$ in
(a) $\left( 2n\pi -\dfrac{3\pi }{4},2n\pi +\dfrac{\pi }{4} \right),n\in Z$
(b) $\left( 2n\pi -\dfrac{\pi }{4},2n\pi +\dfrac{\pi }{6} \right),n\in Z$
(c) $\left( 2n\pi -\dfrac{\pi }{3},2n\pi +\dfrac{\pi }{3} \right),n\in Z$
(d) $\left( 2n\pi -\dfrac{\pi }{4},2n\pi +\dfrac{3\pi }{4} \right),n\in Z$

Answer 1

We start solving the problem by substituting the results $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta$, $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta$ and $\sin 2\theta =2\sin \theta \cos \theta$ in the given equation. We then make the necessary calculations and then make use of the results ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)$, ${{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1$ to proceed through the problem. We then divide the obtained inequality with $\sqrt{2}$ on both sides and make use of the results $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$, $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ to proceed through the problem. We then use of the result that if $\cos x>0$, then the general solution set of x will be $x\in \left( 2n\pi -\dfrac{\pi }{2},2n\pi +\dfrac{\pi }{2} \right)$, $n\in Z$ to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the values of $\theta$ for which the expression $\cos 3\theta +\sin 3\theta +\left( 2\sin 2\theta -3 \right)\left( \sin \theta -\cos \theta \right)$ is positive.
We have $\cos 3\theta +\sin 3\theta +\left( 2\sin 2\theta -3 \right)\left( \sin \theta -\cos \theta \right)>0$ ---(1).
We know that $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta$, $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta$ and $\sin 2\theta =2\sin \theta \cos \theta$. Let us use these results in equation (1).
$\Rightarrow 4{{\cos }^{3}}\theta -3\cos \theta +3\sin \theta -4{{\sin }^{3}}\theta +\left( 4\sin \theta \cos \theta -3 \right)\left( \sin \theta -\cos \theta \right)>0$.
$\Rightarrow 4{{\cos }^{3}}\theta -3\cos \theta +3\sin \theta -4{{\sin }^{3}}\theta +4{{\sin }^{2}}\theta \cos \theta -3\sin \theta -4\sin \theta {{\cos }^{2}}\theta +3\cos \theta >0$.
$\Rightarrow 4{{\cos }^{3}}\theta -4{{\sin }^{3}}\theta +4{{\sin }^{2}}\theta \cos \theta -4\sin \theta {{\cos }^{2}}\theta >0$.
$\Rightarrow {{\cos }^{3}}\theta -{{\sin }^{3}}\theta +{{\sin }^{2}}\theta \cos \theta -\sin \theta {{\cos }^{2}}\theta >0$ ---(2).
We know that ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)$. Let us use this result in equation (2).
$\Rightarrow \left( \cos \theta -\sin \theta \right)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta +\cos \theta \sin \theta \right)-\sin \theta \cos \theta \left( \cos \theta -\sin \theta \right)>0$ ---(3).
We know that ${{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1$. Let us use this result in equation (3).
$\Rightarrow \left( \cos \theta -\sin \theta \right)\left( 1+\cos \theta \sin \theta \right)-\sin \theta \cos \theta \left( \cos \theta -\sin \theta \right)>0$.
$\Rightarrow \left( \cos \theta -\sin \theta \right)\left( 1+\cos \theta \sin \theta -\sin \theta \cos \theta \right)>0$.
$\Rightarrow \left( \cos \theta -\sin \theta \right)>0$ ---(4).
Let us divide both sides of equation (4) with $\sqrt{2}$.
$\Rightarrow \dfrac{1}{\sqrt{2}}\cos \theta -\dfrac{1}{\sqrt{2}}\sin \theta >0$ ---(5).
We know that $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$. Let us use this result in equation (5).
$\Rightarrow \cos \left( \dfrac{\pi }{4} \right)\cos \theta -\sin \left( \dfrac{\pi }{4} \right)\sin \theta >0$ ---(6).
We know that $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$. Let us use this result in equation (6).
$\Rightarrow \cos \left( \dfrac{\pi }{4}+\theta \right)>0$ ---(7).
We know that if $\cos x>0$, then the general solution set of x will be $x\in \left( 2n\pi -\dfrac{\pi }{2},2n\pi +\dfrac{\pi }{2} \right)$, $n\in Z$. Let us use this result in equation (7).
So, we get $\dfrac{\pi }{4}+\theta \in \left( 2n\pi -\dfrac{\pi }{2},2n\pi +\dfrac{\pi }{2} \right)$.
$\Rightarrow \theta \in \left( 2n\pi -\dfrac{\pi }{2}-\dfrac{\pi }{4},2n\pi +\dfrac{\pi }{2}-\dfrac{\pi }{4} \right)$.
$\Rightarrow \theta \in \left( 2n\pi -\dfrac{3\pi }{4},2n\pi +\dfrac{\pi }{4} \right)$, $n\in Z$.
So, we have found the solution set for the given equation $\cos 3\theta +\sin 3\theta +\left( 2\sin 2\theta -3 \right)\left( \sin \theta -\cos \theta \right)$ as $\theta \in \left( 2n\pi -\dfrac{3\pi }{4},2n\pi +\dfrac{\pi }{4} \right)$, $n\in Z$.

So, the correct answer is “Option a”.

Note: We need to perform each step carefully in order to avoid confusion. We should be confused while applying the formulas of $\cos 3\theta$, $\sin 3\theta$ and $\sin 2\theta$ while solving this problem. We should not confuse the general solution set while solving the problems related to trigonometric equations. Similarly, we can expect problems to find the principle solution for the given problem.