Question

How do I solve $2{{\sec }^{2}}x+{{\tan }^{2}}x-3=0?$

Answer 1

Hint: We use the basic trigonometric identities to solve the given problem. We may use the identity that says $\sec x=\dfrac{1}{\cos x}$ which is a reciprocal relation. We use another reciprocal relation $\tan x=\dfrac{\sin x}{\cos x}.$ Also, we use the Pythagorean relation ${{\sin }^{2}}x+{{\cos }^{2}}x=1.$

Complete step by step solution:
Consider the given equation which includes the trigonometric functions,
$\Rightarrow 2{{\sec }^{2}}x+{{\tan }^{2}}x-3=0$
To solve this equation, we use the reciprocal relations which we have learnt already, $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}.$ When we square both of these relations, we get ${{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}$ and ${{\tan }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}.$
Now, we are going to apply these relations in the given equations to get,
$\Rightarrow 2\dfrac{1}{{{\cos }^{2}}x}+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}-3=0$
In the above equation, we have substituted for ${{\sec }^{2}}x$ and ${{\tan }^{2}}x.$
In the next step, we are going to multiply the whole equation with ${{\cos }^{2}}x.$
We will get,
$\Rightarrow \dfrac{2{{\cos }^{2}}x}{{{\cos }^{2}}x}+\dfrac{{{\cos }^{2}}x{{\sin }^{2}}x}{{{\cos }^{2}}x}-3{{\cos }^{2}}x=0$
We are cancelling the common factors from both the numerator and the denominator.
So, we get,
$\Rightarrow 2+{{\sin }^{2}}x-3{{\cos }^{2}}x=0$
We know that $-3{{\cos }^{2}}x={{\cos }^{2}}x-4{{\cos }^{2}}x$
Since we have to simplify the above obtained equation, we use the above fact we have written.
So, we get
$\Rightarrow 2+{{\sin }^{2}}x+{{\cos }^{2}}x-4{{\cos }^{2}}x=0$
Now with what we get, we are supposed to use the Pythagorean relation that connects the sine function and the cosine function ${{\sin }^{2}}x+{{\cos }^{2}}x=1.$
The equation will now become,
$\Rightarrow 2+1-4{{\cos }^{2}}x=0$
We add the numbers as usual to get,
$\Rightarrow 3-4{{\cos }^{2}}x=0$
Now we are transposing $3$ from the left-hand side to the right-hand side,
$\Rightarrow -4{{\cos }^{2}}x=-3$
Now we multiply the above equation with $-1.$
$\Rightarrow 4{{\cos }^{2}}x=3$
Now we are transposing $4$ from the left-hand side to the right-hand side,
$\Rightarrow {{\cos }^{2}}x=\dfrac{3}{4}$
Now we have to find the square root of the whole equation as follows,
$\Rightarrow \cos x=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{\sqrt{4}}=\dfrac{\sqrt{3}}{2}$
So, we have to find the value of $x$ for the value of $\cos x$ becomes $\dfrac{\sqrt{3}}{2}.$
On the other hand, $x={{\cos }^{-1}}\dfrac{\sqrt{3}}{2}.$
We know that $\cos x=\dfrac{\sqrt{3}}{2}$ when $x={{\left( \dfrac{\pi }{3} \right)}^{c}}=30{}^\circ .$
That is, ${{\cos }^{-1}}\dfrac{\sqrt{3}}{2}=\left( \dfrac{\pi }{6} \right){}^\circ =30{}^\circ .$
Therefore, the solution of $2{{\sec }^{2}}x+{{\tan }^{2}}x-3=0$ is $x={{\left( \dfrac{\pi }{6} \right)}^{c}}=30{}^\circ .$

Note:
There is another method to solve this:
$\Rightarrow 2{{\sec }^{2}}x+{{\tan }^{2}}x-3=0$
We use the Pythagorean relation ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ which implies ${{\sec }^{2}}x=1-{{\tan }^{2}}x.$
$\Rightarrow 2\left( 1+{{\tan }^{2}}x \right)+{{\tan }^{2}}x-3=0$
$\Rightarrow 2+2{{\tan }^{2}}x+{{\tan }^{2}}x-3=0$
$\Rightarrow 3{{\tan }^{2}}x-1=0$
$\Rightarrow {{\tan }^{2}}x=\dfrac{1}{3}$
$\Rightarrow \tan x=\dfrac{1}{\sqrt{3}}$
We use the reciprocal relation $\tan x=\dfrac{\sin x}{\cos x},$
$\Rightarrow \dfrac{\sin x}{\cos x}=\dfrac{1}{\sqrt{3}}=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$
We know that $\sin x=\dfrac{1}{2} \text{and} \cos x=\dfrac{\sqrt{3}}{2}\Rightarrow x=\left( \dfrac{\pi }{6} \right).$