Question

Find the general solution for ${{\sec }^{2}}2x=1-\tan 2x$

Answer 1

Hint: First we will use the formula ${{\sec }^{2}}2x=1+{{\tan }^{2}}2x$ , and then we will rearrange the expression such that we take variable to one side and then we take the common terms out to convert into two different equation and then we will use formula for general formula for tan to find the final answer.

Complete step-by-step answer:
Let’s start solving the question,
${{\sec }^{2}}2x=1-\tan 2x$
Now we will use the formula ${{\sec }^{2}}2x=1+{{\tan }^{2}}2x$, to convert the whole equation in tan,
Therefore we get,
$1+{{\tan }^{2}}2x=1-\tan 2x$
$\begin{aligned} & {{\tan }^{2}}2x+\tan 2x=0 \\\ & \tan 2x\left( \tan 2x+1 \right)=0 \\\ \end{aligned}$
Now we have converted it into two equation and we will solve it separately,
We get tan2x = 0 and tan2x + 1 = 0.
Let’s first solve tan2x = 0,
We know that tan0 = 0,
Therefore we can say that tan2x = tan0,
Now we will use the formula for general solution of tan,
Now, if we have $\tan \theta =\tan \alpha$ then the general solution is:
$\theta =n\pi +\alpha$
Now using the above formula for tan2x = tan0 we get,
$\begin{aligned} & 2x=n\pi +0 \\\ & x=\dfrac{n\pi }{2}.............(1) \\\ \end{aligned}$
Here n = integer.
Now we will solve tan2x + 1 = 0,
$\begin{aligned} & \tan 2x=-1 \\\ & \tan 2x=\tan \left( \dfrac{-\pi }{4} \right) \\\ \end{aligned}$
Now we will use the formula for general solution of tan,
Now, if we have $\tan \theta =\tan \alpha$ then the general solution is:
$\theta =n\pi +\alpha$
Now using the above formula for $\tan 2x=\tan \left( \dfrac{-\pi }{4} \right)$ we get,
$\begin{aligned} & 2x=n\pi +\left( \dfrac{-\pi }{4} \right) \\\ & x=\dfrac{n\pi }{2}+\left( \dfrac{-\pi }{8} \right).............(2) \\\ \end{aligned}$
Here n = integer.
Now from equation (1) and (2) we can say that the answer is,
$x=\dfrac{n\pi }{2}\text{ or }x=\dfrac{n\pi }{2}+\left( \dfrac{-\pi }{8} \right)$
Hence, this is the answer to this question.

Note: The trigonometric formula ${{\sec }^{2}}2x=1+{{\tan }^{2}}2x$ that we have used must be kept in mind. One can also take some different value of $\alpha$ like in tan2x = 0 we can take $\pi$ instead of 0 , and then can apply the same formula for the general solution, and the answer that we get will also be correct.