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

Hint: 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:
\[\begin{align}
  & \Rightarrow {{t}^{2}}-\dfrac{4}{3}=0 \\
 & \Rightarrow {{t}^{2}}=\dfrac{4}{3} \\
\end{align}\]
Taking square root of both sides, we get
\[\Rightarrow t=\pm \dfrac{2}{\sqrt{3}}\]