Question
Question: How do you solve \({\sec ^2}x - 2{\tan ^2}x = 0\)?...
How do you solve sec2x−2tan2x=0?
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 x satisfying tan(x)=1 by taking the inverse tan of both sides of the equation to extract x from inside the tan. Next, find the value of x satisfying tan(x)=−1 using trigonometric properties. Then, we will get all solutions of the given equation.
Formula used:
sec2θ=tan2θ+1
tan4π=1
tan(π−x)=−tanx
tan(2π−x)=−tanx
Complete step by step solution:
Given equation: sec2x−2tan2x=0
We have to find all possible values of x satisfying a given equation.
First, use identity sec2θ=tan2θ+1 in the above equation.
tan2x+1−2tan2x=0
⇒tan2x=1
Take the square root on both sides of the equation.
⇒tanx=±1
Now, we will find the values of x satisfying tan(x)=1.
So, take the inverse tan of both sides of the equation to extract x from inside the tan.
x=arctan(1)
Since, the exact value of arctan(1)=4π.
⇒x=4π
Since, the tan function is positive in the first and third quadrants.
So, to find the second solution, add the reference angle from π to find the solution in the fourth quadrant.
x=π+4π
⇒x=45π
Since, the period of the tan(x) function is π so values will repeat every π radians in both directions.
x=4π+nπ,45π+nπ, for any integer n.
Now, we will find the values of x satisfying tan(x)=−1…(i)
So, using the property tan(π−x)=−tanx and tan4π=1 in equation (i).
⇒tan(x)=−tan4π
⇒tan(x)=tan(π−4π)
⇒x=43π
Now, using the property tan(2π−x)=−tanx and tan4π=1 in equation (i).
⇒tan(x)=−tan4π
⇒tan(x)=tan(2π−4π)
⇒x=47π
Since, the period of the tan(x) function is π so values will repeat every π radians in both directions.
x=43π+nπ,47π+nπ, for any integer n.
Final solution: Hence, x=4π+nπ,43π+nπ,45π+nπ,47π+nπ, for any integer n are solutions of the given equation.
Note:
In the above question, we can find the solutions of a given equation by plotting the equation, sec2x−2tan2x=0 on graph paper and determine all its solutions.
From the graph paper, we can see that x=4π, x=43π, x=45π and x=47π are solution of given equation, and solution repeat every π radians in both directions.
So, these will be the solutions of the given equation.
Final solution: Hence, x=4π+nπ,43π+nπ,45π+nπ,47π+nπ, for any integer nare solutions of the given equation.