Question

Find the value of $\theta$ if $\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0$ where ${{0}^{\circ }}\text{ }<\text{ }\theta \text{ }<\text{ }{{180}^{\circ }}$.

Answer 1

In this question, we are given an equation in terms of cosine and we have to find the value of $\theta$ where $\theta$ lies between ${{0}^{\circ }}\text{ }and\text{ }{{180}^{\circ }}$. For this, we will first simplify the given equation (using various trigonometric identities) into factors of cosine. After that, we will find a general solution of cosine when $\cos \theta ={{0}^{\circ }}$. Since we need to find $\theta$ between ${{0}^{\circ }}\text{ }and\text{ }{{180}^{\circ }}$. So we will use general values to find values that lie between ${{0}^{\circ }}\text{ }and\text{ }{{180}^{\circ }}$. Trigonometric formulas and identities that we will use are:

& \left( i \right)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ & \left( ii \right)1+\cos 2\theta =2{{\cos }^{2}}\theta \\\ & \left( iii \right)\cos \theta =0\Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{2}\text{ where }n=0,1,2\cdots \\\ \end{aligned}$$ **Complete step by step answer:** Here, we are given equation as, $$\cos 6\theta +\cos 4\theta +\cos 2\theta +1=0$$ Let us first rearrange the left side, we get: $$\left( \cos 6\theta +\cos 2\theta \right)+\left( \cos 4\theta +1 \right)=0$$ As we know that, $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ so applying it on $\left( \cos 6\theta +\cos 2\theta \right)$ we get: $$\begin{aligned} & 2\cos \left( \dfrac{6+2}{2} \right)\theta \cos \left( \dfrac{6-2}{2} \right)\theta +\left( \cos 4\theta +1 \right)=0 \\\ & \Rightarrow 2\cos 4\theta \cos 2\theta +\cos 4\theta +1=0 \\\ \end{aligned}$$ As we know that, $1+\cos 2\theta =2{{\cos }^{2}}\theta $ so applying it on $\cos 4\theta +1$ we get: $$\Rightarrow 2\cos 4\theta \cos 2\theta +2{{\cos }^{2}}2\theta =0$$ Taking $2\cos 2\theta $ from left side of the equation, we get: $$\Rightarrow 2\cos 2\theta \left( \cos 4\theta +\cos 2\theta \right)=0$$ Applying $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ on $\cos 4\theta +\cos 2\theta $ we get: $$\begin{aligned} & \Rightarrow 2\cos 2\theta \left( 2\cos \left( \dfrac{4+2}{2} \right)\theta \cos \left( \dfrac{4-2}{2} \right)\theta \right)=0 \\\ & \Rightarrow 2\cos 2\theta \left( 2\cos 3\theta \cos \theta \right)=0 \\\ & \Rightarrow 4\cos 2\theta \cos 3\theta \cos \theta =0 \\\ \end{aligned}$$ Hence, we have factorized the given equation. Now, we know that if $a\times b\times c=0$ then a = 0 or b = 0 or c = 0. Similarly, we get: $$\cos 2\theta ,\cos 3\theta \text{ and }\cos \theta =0$$ Since, $\cos \theta =0\Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{2}\text{ where }n=0,1,2\cdots $ so let us apply this on all above factors, we get: For $\cos 2\theta =0$, $$\begin{aligned} & 2\theta =\left( 2n+1 \right)\dfrac{\pi }{2}\text{ where }n=0,1,2,3\cdots \\\ & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{4}\text{ where }n=0,1,2,3\cdots \\\ \end{aligned}$$ For $\cos 3\theta =0$, $$\begin{aligned} & 3\theta =\left( 2k+1 \right)\dfrac{\pi }{2}\text{ where }k=0,1,2,3\cdots \\\ & \Rightarrow \theta =\left( 2k+1 \right)\dfrac{\pi }{6}\text{ where }k=0,1,2,3\cdots \\\ \end{aligned}$$ For $\cos \theta =0$, $$\theta =\left( 2p+1 \right)\dfrac{\pi }{2}\text{ where }p=0,1,2,3\cdots $$ But we need values of $\theta $ only between ${{0}^{\circ }}\text{ }and\text{ }{{180}^{\circ }}$ i.e. between ${{0}^{\circ }}\text{ }and\text{ }\pi $ therefore putting n = 0, 1 we get: $\theta =\dfrac{\pi }{4}\text{ and }\theta =\dfrac{3\pi }{4}$. Putting k = 0, 1, 2 we get: $\theta =\dfrac{\pi }{6}\text{,}\theta =\dfrac{\pi }{2}\text{ and }\theta =\dfrac{5\pi }{6}$. Putting p = 0 we get: $\theta =\dfrac{\pi }{2}$. **Hence, required value of $\theta $ are $\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{5\pi }{6}$. In degree value of $\theta $ are ${{45}^{\circ }},{{135}^{\circ }},{{30}^{\circ }},{{90}^{\circ }}\text{ and }{{150}^{\circ }}$.** **Note:** Students should carefully find value of $\theta $ between ${{0}^{\circ }}\text{ }and\text{ }{{180}^{\circ }}$ from general values of $\theta $. Student should not write repeated values again in final answer. While applying the trigonometric formula $\cos C+\cos D$ take care of the signs. Always remember that in $\left( 2n+1 \right)\dfrac{\pi }{2}$ n starts from 0 and not from 1. n = 0 should always be considered. Students should keep in mind all the trigonometric identities and formulas for solving these types of sums.