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Question: How do you solve \({\cos ^2}(3x) = 1?\)...

How do you solve cos⁡2(3x)=1?{\cos ^2}(3x) = 1?

Explanation

Solution

In order to solve this trigonometric question, you should have knowledge about the general solution for cos⁡θ=±1\cos \theta = \pm 1 , in this type of problems you always have to solve for the general solution of the given trigonometric equation.
General solution for cos⁡θ=±1\cos \theta = \pm 1 is given as θ=kπ,  where  k∈I\theta = k\pi ,\;{\text{where}}\;k \in I

Complete step by step solution:
To solve for cos⁡2(3x)=1{\cos ^2}(3x) = 1 we should first consider the argument of the cosine function to be θ\theta in order to make the process easy and understandable.
⇒θ=3x\Rightarrow \theta = 3x
So we can rewrite the given trigonometric equation as follows
⇒cos⁡2(3x)=1 ⇒cos⁡2θ=1  \Rightarrow {\cos ^2}(3x) = 1 \\\ \Rightarrow {\cos ^2}\theta = 1 \\\
Solving it further we will get,
⇒cos⁡2θ=1 ⇒cos⁡θ=±1  \Rightarrow {\cos ^2}\theta = 1 \\\ \Rightarrow \cos \theta = \pm 1 \\\
Now we are all familiar with the general solution of cos⁡θ=1  and  cos⁡θ=−1\cos \theta = 1\;{\text{and}}\;\cos \theta = - 1,
If you are not then let us understand first what is a general solution in trigonometry.
Since all the trigonometric functions are periodic in nature, that is all of them repeat their values after a fixed interval of angles or you say argument.
So they will definitely have an infinite number of solutions for a particular value. Here the general solution comes: it is the complete set of values of the unknown arguments or angles satisfying the equation.
Now the general solution for cos⁡θ=1  and  cos⁡θ=−1\cos \theta = 1\;{\text{and}}\;\cos \theta = - 1 are respectively
θ=2kπ  and  θ=(2k+1)π,  where  k∈I\theta = 2k\pi \;and\;\theta = (2k + 1)\pi ,\;{\text{where}}\;k \in I
From this, we can write the general solution for cos⁡θ=±1\cos \theta = \pm 1 as
θ=kπ,  where  k∈I\theta = k\pi ,\;{\text{where}}\;k \in I
Therefore we can solve further as
⇒cos⁡θ=±1 ⇒θ=kπ,  where  k∈I  \Rightarrow \cos \theta = \pm 1 \\\ \Rightarrow \theta = k\pi ,\;{\text{where}}\;k \in I \\\
Putting back θ=3x\theta = 3x we will get
⇒θ=kπ,  where  k∈I ⇒3x=kπ,  where  k∈I ⇒x=kπ3,  where  k∈I  \Rightarrow \theta = k\pi ,\;{\text{where}}\;k \in I \\\ \Rightarrow 3x = k\pi ,\;{\text{where}}\;k \in I \\\ \Rightarrow x = \dfrac{{k\pi }}{3},\;{\text{where}}\;k \in I \\\
Therefore the general solution for cos⁡θ=±1\cos \theta = \pm 1 is x=kπ3,  where  k∈Ix = \dfrac{{k\pi }}{3},\;{\text{where}}\;k \in I

Note: There are three types of solution:
1. Principal solution: It is the smallest value of the unknown angle satisfying the equation.
2. Particular solution: A specific value satisfying the equation.
3. General solution: It is a complete set of values of unknown angles satisfying the equation.
Normally the general solution is preferred.