Question

Solve $\cos 3x+\cos 2x=\sin \left( \dfrac{3x}{2} \right)+\sin \left( \dfrac{x}{2} \right),0\le x<2\pi$.

Answer 1

Hint: We will first start by collecting the terms of cosine on one side and sine on other terms. Then we will use the formula that,

& \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ \end{aligned}$$ and simplify to find the answer. Complete step-by-step answer: Now, we have been given that the value of $\cos 3x+\cos 2x=\sin \left( \dfrac{3x}{2} \right)+\sin \left( \dfrac{x}{2} \right)$ for $0\le x<2\pi $. Now, we know the trigonometric identity that, $$\begin{aligned} & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ \end{aligned}$$ So, we have the equation as, $\begin{aligned} & 2\cos \left( \dfrac{5x}{2} \right)\cos \left( \dfrac{x}{2} \right)=2\sin \left( \dfrac{4x}{4} \right)\cos \left( \dfrac{x}{2} \right) \\\ & \cos \left( \dfrac{5x}{2} \right)\cos \left( \dfrac{x}{2} \right)=\sin \left( x \right)\cos \left( \dfrac{x}{2} \right) \\\ \end{aligned}$ Now, we can rewrite it as, $\cos \left( \dfrac{x}{2} \right)\cos \left( \dfrac{5x}{2} \right)-\cos \left( \dfrac{x}{2} \right)\sin \left( x \right)=0$ Now, we will take $\cos \dfrac{x}{2}$ common in equation. So, we have, $\cos \left( \dfrac{x}{2} \right)\left\\{ \cos \left( \dfrac{5x}{2} \right)-\sin \left( x \right) \right\\}=0$ Now, we have either $\cos \left( \dfrac{x}{2} \right)=0\ or\ \cos \left( \dfrac{5x}{2} \right)-\sin \left( x \right)=0$. Now, we know the trigonometric identity that, $\begin{aligned} & \sin \left( x \right)=\cos \left( \dfrac{\pi }{2}-x \right) \\\ & \Rightarrow \cos \left( \dfrac{x}{2} \right)=0 \\\ & \cos \left( \dfrac{5x}{2} \right)-\cos \left( \dfrac{\pi }{2}-x \right)=0 \\\ \end{aligned}$ Now, we know that if $\cos x=\cos \theta $ then $x=2n\pi \pm \theta $. Therefore, we have the values as, $\cos \left( \dfrac{x}{2} \right)=0\ or\ \cos \dfrac{5x}{2}=\cos \left( \dfrac{\pi }{2}-x \right)$ Now, since we know that $\cos \dfrac{\pi }{2}=0$. Therefore, $\begin{aligned} & \cos \dfrac{x}{2}=\cos \dfrac{\pi }{2} \\\ & \dfrac{x}{2}=2n\pi \pm \dfrac{\pi }{2}\ for\ n\in \mathcal{Z}............\left( 1 \right) \\\ \end{aligned}$ Similarly, we have, $\dfrac{5x}{2}=2n\pi \pm \left( \dfrac{\pi }{2}-x \right)\ for\ n\in N.........\left( 2 \right)$ We will taking positive sign, $\begin{aligned} & \dfrac{5x}{2}=2n\pi +\dfrac{\pi }{2}-x \\\ & \dfrac{5x}{2}+x=2n\pi +\dfrac{\pi }{2} \\\ & \dfrac{5x+2x}{2}=\dfrac{4n\pi +\pi }{2} \\\ & \dfrac{7x}{2}=\dfrac{\pi \left( 4n+1 \right)}{2} \\\ & 7x=\pi \left( 4n+1 \right) \\\ & x=\dfrac{\left( 4n+1 \right)\pi }{7} \\\ \end{aligned}$ Now, for $n=0,1,2,3$we have, $$x=\dfrac{\pi }{7},\dfrac{5\pi }{7},\dfrac{9\pi }{7},\dfrac{13\pi }{7}$$ Now, if we take negative in (2). So, we have, $\begin{aligned} & \dfrac{5x}{2}=2n\pi -\dfrac{\pi }{2}+x \\\ & \dfrac{5x}{2}-x=\dfrac{\pi \left( 4n-1 \right)}{2} \\\ & \dfrac{5x-2x}{2}=\dfrac{\pi \left( 4n-1 \right)}{2} \\\ & \dfrac{3x}{2}=\dfrac{\pi \left( 4n-1 \right)}{2} \\\ & x=\dfrac{\pi }{3}\left( 4n-1 \right) \\\ \end{aligned}$ Now, if we put n = 1, we have, $x=\dfrac{\pi }{3}\left( 3 \right)=\pi $ Now, from (1) we have, $\dfrac{x}{2}=2n\pi \pm \dfrac{\pi }{2}$ Now, if we put n = 0, so we have, $\begin{aligned} & \dfrac{x}{2}=\pm \dfrac{\pi }{2} \\\ & x=\pm \pi \\\ \end{aligned}$ Now, since we have given $0\le x<2\pi $. Therefore, x can’t be negative. Hence, the value of x is $\pi $. Now, total values of x are, $$\dfrac{\pi }{7},\dfrac{5\pi }{7},\dfrac{9\pi }{7},\dfrac{13\pi }{7},\pi $$ Note: It is important to note that there are infinitely many solutions for $\sin x=\sin \theta $. But since we have been given that $0\le x<2\pi $ therefore, we have to limit the values in this range out of all the values.