Question

How do you find all solutions of the equation below in the interval $\left[ 0,2\pi \right)$ for $9{{\sec }^{2}}x-12=0$?

Answer 1

To solve the following problem, we will need to use the substitution method. We will substitute a variable t for sec(x) in the given equation. By doing this, we will get a quadratic equation in t, by solving this equation we will get the solution value for t or sec(x). After this we will use the inverse trigonometric functions. For this question, we should know that the algebraic expressions of the form ${{x}^{2}}-{{a}^{2}}$ can be simplified as ${{x}^{2}}-{{a}^{2}}=\left( x-a \right)\left( x+a \right)$.

Complete step by step solution:
We are asked to solve the equation $9{{\sec }^{2}}x-12=0$. Dividing both sides of this equation by 9, we get
$\Rightarrow \dfrac{9{{\sec }^{2}}x}{9}-\dfrac{12}{9}=0$
Cancelling out the common factors, we get
$\Rightarrow {{\sec }^{2}}x-\dfrac{4}{3}=0$
Put t at the place sec(x) in above equation, thus we get an equation in terms of t as
$\Rightarrow {{t}^{2}}-\dfrac{4}{3}=0$
We can express this equation as
$\Rightarrow {{t}^{2}}-{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}=0$
Using the algebraic expansion ${{x}^{2}}-{{a}^{2}}=\left( x-a \right)\left( x+a \right)$, we can simplify the equation as
$\Rightarrow \left( t-\dfrac{2}{\sqrt{3}} \right)\left( t+\dfrac{2}{\sqrt{3}} \right)=0$
From the above equation, we get the roots as $t=-\dfrac{2}{\sqrt{3}}\And t=\dfrac{2}{\sqrt{3}}$.
Using the previous substitution, we get $\sec x=-\dfrac{2}{\sqrt{3}}\And \sec x=\dfrac{2}{\sqrt{3}}$.
First let’s find the solution for $\sec x=-\dfrac{2}{\sqrt{3}}$in the range of $\left[ 0,2\pi \right)$. There are only two values which satisfy this, $x=\dfrac{5\pi }{6}\And x=\dfrac{7\pi }{6}$. Similarly, for $\sec x=\dfrac{2}{\sqrt{3}}$also there are only two values $x=\dfrac{\pi }{6}\And x=\dfrac{11\pi }{6}$. Thus, the given equation has following solutions $x=\dfrac{5\pi }{6},x=\dfrac{7\pi }{6},x=\dfrac{\pi }{6}\And x=\dfrac{11\pi }{6}$.

Note: We can also use cosine ratio as we are more comfortable with it. As secant and cosine are inverse of each other, we get $cosx=-\dfrac{\sqrt{3}}{2}\And cosx=\dfrac{\sqrt{3}}{2}$. This will also give the same solution we get above.
For equation ${{t}^{2}}-\dfrac{4}{3}=0$, we can also solve it without using the algebraic expression as follows:

& \Rightarrow {{t}^{2}}-\dfrac{4}{3}=0 \\\ & \Rightarrow {{t}^{2}}=\dfrac{4}{3} \\\ \end{aligned}$$ Taking square root of both sides, we get $$\Rightarrow t=\pm \dfrac{2}{\sqrt{3}}$$