Question

Solve the following trigonometric equation:
$\cos 5x+\cos 3x=0$.

Answer 1

Hint: Use the formula, $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$ to evaluate the given expression, then equate the product of all the terms equal to 0 to find the general solution of the given trigonometric equation.
Complete step-by-step answer:
Trigonometric equations are equations involving trigonometric functions. A trigonometric equation that holds true for any angle is called a trigonometric identity. There are other equations that are true for certain angles. They are generally known as conditional equations. Solution of trigonometric equations requires concepts of different trigonometric identities like conversion from product to sum of trigonometric functions using product to sum rule and similarly conversion from sum to product using sum to product rule.
Now, when we talk about the solutions of trigonometric functions, there are two types of solutions: general solution and principal solution. There are an infinite number of positive and negative angles that satisfy an equation. We cannot write all of them, so we generalize them using an integer so that one can find any solution by just putting the value of integer. However, in principle there is a limit given to us between which we are required to find the solution. Here we are going to use the sum to product rule.
Now, let us come to the question. We have been given, $\cos 5x+\cos 3x=0$. Applying the formula, $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$, we get,

& \cos 5x+\cos 3x=0 \\\ & \Rightarrow 2\cos \left( \dfrac{5x+3x}{2} \right)\cos \left( \dfrac{5x-3x}{2} \right)=0 \\\ & \Rightarrow 2\cos 4x\cos x=0 \\\ \end{aligned}$$ Substituting each term equal to 0 we get, $\cos 4x=0\text{ or }\cos x=0.................(i)$ We know that the general solution of $\cos y=0$ is given by, $y=\left( 2n+1 \right)\times \dfrac{\pi }{2}$. Therefore the solution of equation (i) is: $\begin{aligned} & 4x=(2n+1)\dfrac{\pi }{2}\text{ or }x=(2n+1)\dfrac{\pi }{2} \\\ & \Rightarrow x=(2n+1)\dfrac{\pi }{8}\text{ or }x=(2n+1)\dfrac{\pi }{2} \\\ \end{aligned}$ Here, $n\text{ is any integer}\text{.}$ Note: It is important to note that if, $\cos y=\cos a\Rightarrow y=2n\pi \pm a$. Remember that we have to put each multiplication term equal to zero to find the general solution. We know that the cosine of an odd multiple of $\dfrac{\pi }{2}$ is zero. In the above solution $\left( 2n+1 \right)$ is an odd number.