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Question: How do you solve \({\sec ^2}x - 2{\tan ^2}x = 0\)?...

How do you solve sec⁡2x−2tan⁡2x=0{\sec ^2}x - 2{\tan ^2}x = 0?

Explanation

Solution

First, we have to simplify the given equation using trigonometry identities. Then, take the square root on both sides of the equation. Next, find the values of xx satisfying tan⁡(x)=1\tan \left( x \right) = 1 by taking the inverse tan of both sides of the equation to extract xx from inside the tan. Next, find the value of xx satisfying tan⁡(x)=−1\tan \left( x \right) = - 1 using trigonometric properties. Then, we will get all solutions of the given equation.
Formula used:
sec⁡2θ=tan⁡2θ+1{\sec ^2}\theta = {\tan ^2}\theta + 1
tan⁡π4=1\tan \dfrac{\pi }{4} = 1
tan⁡(π−x)=−tan⁡x\tan \left( {\pi - x} \right) = - \tan x
tan⁡(2π−x)=−tan⁡x\tan \left( {2\pi - x} \right) = - \tan x

Complete step by step solution:
Given equation: sec⁡2x−2tan⁡2x=0{\sec ^2}x - 2{\tan ^2}x = 0
We have to find all possible values of xx satisfying a given equation.
First, use identity sec⁡2θ=tan⁡2θ+1{\sec ^2}\theta = {\tan ^2}\theta + 1 in the above equation.
tan⁡2x+1−2tan⁡2x=0{\tan ^2}x + 1 - 2{\tan ^2}x = 0
⇒tan⁡2x=1\Rightarrow {\tan ^2}x = 1
Take the square root on both sides of the equation.
⇒tan⁡x=±1\Rightarrow \tan x = \pm 1
Now, we will find the values of xx satisfying tan⁡(x)=1\tan \left( x \right) = 1.
So, take the inverse tan of both sides of the equation to extract xx from inside the tan.
x=arctan⁡(1)x = \arctan \left( 1 \right)
Since, the exact value of arctan⁡(1)=π4\arctan \left( 1 \right) = \dfrac{\pi }{4}.
⇒x=π4\Rightarrow x = \dfrac{\pi }{4}
Since, the tan function is positive in the first and third quadrants.
So, to find the second solution, add the reference angle from π\pi to find the solution in the fourth quadrant.
x=π+π4x = \pi + \dfrac{\pi }{4}
⇒x=5π4\Rightarrow x = \dfrac{{5\pi }}{4}
Since, the period of the tan⁡(x)\tan \left( x \right) function is π\pi so values will repeat every π\pi radians in both directions.
x=π4+nπ,5π4+nπx = \dfrac{\pi }{4} + n\pi ,\dfrac{{5\pi }}{4} + n\pi , for any integer nn.
Now, we will find the values of xx satisfying tan⁡(x)=−1\tan \left( x \right) = - 1…(i)
So, using the property tan⁡(π−x)=−tan⁡x\tan \left( {\pi - x} \right) = - \tan x and tan⁡π4=1\tan \dfrac{\pi }{4} = 1 in equation (i).
⇒tan⁡(x)=−tan⁡π4\Rightarrow \tan \left( x \right) = - \tan \dfrac{\pi }{4}
⇒tan⁡(x)=tan⁡(π−π4)\Rightarrow \tan \left( x \right) = \tan \left( {\pi - \dfrac{\pi }{4}} \right)
⇒x=3π4\Rightarrow x = \dfrac{{3\pi }}{4}
Now, using the property tan⁡(2π−x)=−tan⁡x\tan \left( {2\pi - x} \right) = - \tan x and tan⁡π4=1\tan \dfrac{\pi }{4} = 1 in equation (i).
⇒tan⁡(x)=−tan⁡π4\Rightarrow \tan \left( x \right) = - \tan \dfrac{\pi }{4}
⇒tan⁡(x)=tan⁡(2π−π4)\Rightarrow \tan \left( x \right) = \tan \left( {2\pi - \dfrac{\pi }{4}} \right)
⇒x=7π4\Rightarrow x = \dfrac{{7\pi }}{4}
Since, the period of the tan⁡(x)\tan \left( x \right) function is π\pi so values will repeat every π\pi radians in both directions.
x=3π4+nπ,7π4+nπx = \dfrac{{3\pi }}{4} + n\pi ,\dfrac{{7\pi }}{4} + n\pi , for any integer nn.
Final solution: Hence, x=π4+nπ,3π4+nπ,5π4+nπ,7π4+nπx = \dfrac{\pi }{4} + n\pi ,\dfrac{{3\pi }}{4} + n\pi ,\dfrac{{5\pi }}{4} + n\pi ,\dfrac{{7\pi }}{4} + n\pi , for any integer nn are solutions of the given equation.

Note:
In the above question, we can find the solutions of a given equation by plotting the equation, sec⁡2x−2tan⁡2x=0{\sec ^2}x - 2{\tan ^2}x = 0 on graph paper and determine all its solutions.

From the graph paper, we can see that x=π4x = \dfrac{\pi }{4}, x=3π4x = \dfrac{{3\pi }}{4}, x=5π4x = \dfrac{{5\pi }}{4} and x=7π4x = \dfrac{{7\pi }}{4} are solution of given equation, and solution repeat every π\pi radians in both directions.
So, these will be the solutions of the given equation.
Final solution: Hence, x=π4+nπ,3π4+nπ,5π4+nπ,7π4+nπx = \dfrac{\pi }{4} + n\pi ,\dfrac{{3\pi }}{4} + n\pi ,\dfrac{{5\pi }}{4} + n\pi ,\dfrac{{7\pi }}{4} + n\pi , for any integer nnare solutions of the given equation.