Question

How do you solve $2{{\tan }^{2}}x+{{\sec }^{2}}x=2$ ?

Answer 1

We have been given an equation in trigonometric functions, tangent and secant of x. In order to solve the given equation, we shall first convert the entire function in terms of only one trigonometric function, that is, in terms of only tangent function or only secant function. Then. We shall simplify the equation, find its square root and further obtain the solution set for the trigonometric function.

Complete step by step solution:
Given that $2{{\tan }^{2}}x+{{\sec }^{2}}x=2$.
We know that the according to a trigonometric identity the square of tangent function subtracted from the square of secant function is equal to 1, that is, ${{\sec }^{2}}x-{{\tan }^{2}}x=1$.
From this, we shall substitute the value of the square of secant function as ${{\sec }^{2}}x=1+{{\tan }^{2}}x$.
$\Rightarrow 2{{\tan }^{2}}x+\left( 1+{{\tan }^{2}}x \right)=2$
$\Rightarrow 2{{\tan }^{2}}x+1+{{\tan }^{2}}x=2$
We will transpose the constant term, 1 from left hand side to right hand side of the equation.
$\Rightarrow 2{{\tan }^{2}}x+{{\tan }^{2}}x=2-1$
$\Rightarrow 3{{\tan }^{2}}x=1$
Dividing both sides of the equation by 3, we get
$\Rightarrow {{\tan }^{2}}x=\dfrac{1}{3}$
Now, we will square root both the sides of the equation.
$\Rightarrow \sqrt{{{\tan }^{2}}x}=\sqrt{\dfrac{1}{3}}$
$\Rightarrow \tan x=\dfrac{1}{\sqrt{3}}$
The principle solution of this equation is $\dfrac{\pi }{6},\dfrac{7\pi }{6}$.
However, the general solution of $\tan x=\dfrac{1}{\sqrt{3}}$ is $\left( n\pi +\dfrac{\pi }{6} \right)$ where n is any integer.

Therefore, the solution of the equation, $2{{\tan }^{2}}x+{{\sec }^{2}}x=2$ is $x=n\pi +\dfrac{\pi }{6}$ where n is any integer.

Note: In order to easily determine the values of various trigonometric functions easily, we must have prior knowledge of the graphs of these functions. The principle solution values of any trigonometric function are the ones which are lying in the interval $\left[ 0,2\pi \right)$ only whereas the general solution values are the ones which can occur anywhere in the interval $\left( -\infty ,\infty \right)$.