Question

Solve the equation $\cos \theta +cos3\theta -\cos 2\theta =0$.

Answer 1

Hint: We will apply the trigonometric formula given by $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\cos 3\theta =4co{{s}^{3}}\theta -3\cos \theta$ where $2\theta$ and $3\theta$ are twice and thrice of the angle $\theta$. After applying these, the given equation will be simplified as an equation of degree 3, which can then be solved further.

Complete step-by-step answer:

Consider equation, $\cos \theta +cos3\theta -\cos 2\theta =0.....(1)$
Now we will apply the formula of $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\cos 3\theta =4co{{s}^{3}}\theta -3\cos \theta$ in equation (1).

& \Rightarrow \cos \theta +\left( 4co{{s}^{3}}\theta -3\cos \theta \right)-\left( 2{{\cos }^{2}}\theta -1 \right)=0 \\\ & \Rightarrow \cos \theta +4{{\cos }^{3}}\theta -3\cos \theta -2{{\cos }^{2}}\theta +1=0 \\\ \end{aligned}$$ Now we will arrange the expression in the order of degree of $$cos\theta $$. That is, $$\begin{aligned} & \Rightarrow 4{{\cos }^{3}}\theta -2{{\cos }^{2}}\theta -3\cos \theta +\cos \theta +1=0 \\\ & \Rightarrow 4{{\cos }^{3}}\theta -2{{\cos }^{2}}\theta -2\cos \theta +1=0...(2) \\\ \end{aligned}$$ Now we will put $$x=\cos \theta $$ in equation (2). This will result into $${{x}^{2}}={{\cos }^{2}}\theta $$ and $${{x}^{3}}={{\cos }^{3}}\theta $$. So we get a new expression, given by, $$\begin{aligned} & 4{{\cos }^{3}}\theta -2{{\cos }^{2}}\theta +1=0 \\\ & 4{{x}^{3}}-2{{x}^{2}}-2x+1=0....(3) \\\ \end{aligned}$$ Now we will use the hit and trial method and substitute the value of x in equation (3). First substitute x = 0 in equation (3). $$4{{\left( 0 \right)}^{3}}-2{{\left( 0 \right)}^{2}}-2\left( 0 \right)+1=1$$ This value of x = 0 does not satisfy the equation (3) as the equation does not result into 0. Now we will apply the value $$x=\dfrac{1}{2}$$ and substitute it in equation (3). $$\Rightarrow 4{{\left( \dfrac{1}{2} \right)}^{3}}-2{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{1}{2} \right)+1=\dfrac{4}{8}-\dfrac{2}{4}-1+1=0$$ This clearly implies that $$x=\dfrac{1}{2}$$ is a factor of equation (3) or $$\left( x-\dfrac{1}{2} \right)$$ is a factor. Now we will divide $$x-\dfrac{1}{2}$$ to equation (3). Thus we get, $$x-\dfrac{1}{2}\overset{4{{x}^{2}}-2}{\overline{\left){\begin{aligned} & 4{{x}^{3}}-2{{x}^{2}}-2x+1 \\\ & \underline{4{{x}^{3}}-2{{x}^{2}}} \\\ & -2x+1 \\\ & \underline{-2x+1} \\\ & \underline{000000} \\\ \end{aligned}}\right.}}$$ Clearly we factorize the equation $$4{{x}^{3}}-2{{x}^{2}}-2x+1$$ into $$\left( x-\dfrac{1}{2} \right)\left( 4{{x}^{2}}-2 \right)$$. $$\left( x-\dfrac{1}{2} \right)\left( 4{{x}^{2}}-2 \right)=0$$ Now $$x-\dfrac{1}{2}=0$$ or $$4{{x}^{2}}-2=0$$. Consider $$x-\dfrac{1}{2}=0$$ or $$x=\dfrac{1}{2}$$ since we know $$x=\cos \theta $$. $$\Rightarrow \cos \theta =\dfrac{1}{2}$$. We know that the $$\dfrac{1}{2}$$ is the value of $$\cos \dfrac{\pi }{3}$$. $$\Rightarrow \cos \theta =\dfrac{\pi }{3}$$. The value of cos is positive in the first and fourth quadrant. In quadrant I, we have $$\cos \theta =\dfrac{\pi }{3}$$. $$\begin{aligned} & \Rightarrow \cos \theta =\dfrac{\pi }{3} \\\ & \Rightarrow \theta =\dfrac{\pi }{3} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{\pi }{3}+2n\pi $$, where $$n=0,\pm 1,\pm 2....$$ In fourth quadrant, we have, $$\begin{aligned} & \cos \dfrac{\pi }{3}=\cos \left( 2\pi -\dfrac{\pi }{3} \right) \\\ & \cos \theta =\cos \left( \dfrac{6\pi -\pi }{3} \right) \\\ & \cos \theta =\cos \left( \dfrac{5\pi }{3} \right) \\\ & \theta =\dfrac{5\pi }{3} \\\ \end{aligned}$$ $$\theta =\dfrac{5\pi }{3}+2n\pi $$, where $$n=0,\pm 1,\pm 2....$$ Now we will consider $$4{{x}^{2}}-2=0$$. $$\begin{aligned} & 4{{x}^{2}}=2 \\\ & {{x}^{2}}=\dfrac{1}{2} \\\ & x=\pm \dfrac{1}{\sqrt{2}} \\\ \end{aligned}$$ Since we know that $$\cos \theta =x$$. Thus we can write $$\cos \theta =\dfrac{1}{\sqrt{2}}$$ and $$\cos \theta =\dfrac{-1}{\sqrt{2}}$$. Now we will consider, $$\cos \theta =\dfrac{1}{\sqrt{2}}$$. As we know that $$\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$$. $$\Rightarrow \cos \theta =\cos \dfrac{\pi }{4}$$ The value of cos is positive in the first and fourth quadrant. So we first consider quadrant I. $$\begin{aligned} & \Rightarrow \cos \theta =\cos \dfrac{\pi }{4} \\\ & \Rightarrow \theta =\dfrac{\pi }{4} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{\pi }{4}+2n\pi $$, where $$n=0,\pm 1,\pm 2....$$ Now we will consider the fourth quadrant. In the fourth quadrant, the value of $$\cos \dfrac{\pi }{4}=\cos \left( 2\pi -\dfrac{\pi }{4} \right)$$. $$\begin{aligned} & \Rightarrow \cos \theta =\cos \left( 2\pi -\dfrac{\pi }{4} \right) \\\ & \Rightarrow \cos \theta =\cos \left( \dfrac{8\pi -\pi }{4} \right) \\\ & \Rightarrow \cos \theta =\cos \left( \dfrac{7\pi }{4} \right) \\\ & \Rightarrow \theta =\dfrac{7\pi }{4} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{7\pi }{4}+2n\pi $$, where $$n=0,\pm 1,\pm 2....$$ Now we will consider $$\cos \theta =\dfrac{-1}{\sqrt{2}}$$ as the value of $$\dfrac{1}{\sqrt{2}}=\cos \dfrac{\pi }{4}$$. $$\Rightarrow \cos \theta =-\cos \dfrac{\pi }{4}$$ We know that the value of cos is negative in the second and third quadrant. So we will consider the second quadrant. In this quadrant, the value of $$-\cos \dfrac{\pi }{4}=\cos \left( \pi -\dfrac{\pi }{4} \right)$$. $$\begin{aligned} & \Rightarrow cos\theta =cos\left( \pi -\dfrac{\pi }{4} \right) \\\ & \Rightarrow cos\theta =cos\left( \dfrac{3\pi }{4} \right) \\\ & \Rightarrow \theta =\dfrac{3\pi }{4} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{3\pi }{4}+2n\pi $$, where $$n=0,\pm 1,\pm 2....$$ Also, in the third quadrant the value of $$-\cos \dfrac{\pi }{4}=\cos \left( \pi +\dfrac{\pi }{4} \right)$$. $$\begin{aligned} & \Rightarrow \cos \theta =\cos \left( \pi +\dfrac{\pi }{4} \right) \\\ & \Rightarrow \cos \theta =\cos \left( \dfrac{5\pi }{4} \right) \\\ & \Rightarrow \theta =\dfrac{5\pi }{4} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{5\pi }{4}+2n\pi $$, where $$n=0,\pm 1,\pm 2....$$ Hence the solution of equation (1) is given by, $$\theta =\dfrac{\pi }{3},\dfrac{5\pi }{3},\dfrac{\pi }{4},\dfrac{7\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4}.$$ Note: Alternatively, we could have factorized the equation $$4{{x}^{2}}-2$$ by factorization Actually, we could have used factorization method in the equation $$4{{x}^{3}}-2{{x}^{2}}-2x+1$$. Since we are asked about the basic solution instead of the general solutions that why we write $$\theta =\dfrac{\pi }{2},\theta =\dfrac{2\pi }{3}$$ and so on instead of $$\theta =\dfrac{\pi }{2}+2n\pi ,\theta =\dfrac{2\pi }{3}+2n\pi $$.