Question

How do you solve \[{{\sec }^{2}}(x)-\sec (x)=2\]?

Answer 1

Hint: To solve the given question, we will need some of the following properties. The first is for a quadratic equation \[a{{x}^{2}}+bx+c=0\], using the formula method the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Also, we should know that the general solution of \[x={{\sec }^{-1}}(a)\], here a is a real number in the range of \[\left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)\] is \[2n\pi \pm \theta \]. Where \[\theta \] is the solution in the principal range.

Complete step by step answer:
The given equation is \[{{\sec }^{2}}(x)-\sec (x)=2\]. To make the equation simpler, let’s substitute \[\sec (x)=t\]. By this the given equation becomes,
\[{{t}^{2}}-t=2\]
Subtracting 2 from both sides of the above equation, we get
\[\Rightarrow {{t}^{2}}-t-2=2-2\]
\[\Rightarrow {{t}^{2}}-t-2=0\]
The above equation is quadratic in t. we know that for a quadratic equation \[a{{x}^{2}}+bx+c=0\], using the formula method the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Using this method, we get the roots of the equation \[{{t}^{2}}-t-2=0\] as follows,
\[\begin{align}
  & t=\dfrac{-(-1)\pm \sqrt{{{(-1)}^{2}}-4(1)(-2)}}{2(1)} \\
 & \Rightarrow t=\dfrac{1\pm \sqrt{1+8}}{2}=\dfrac{1\pm \sqrt{9}}{2} \\
 & \Rightarrow t=\dfrac{1\pm \sqrt{9}}{2}=\dfrac{1\pm 3}{2} \\
\end{align}\]
\[\Rightarrow t=\dfrac{1+3}{2}\] or \[t=\dfrac{1-3}{2}\]
\[\Rightarrow t=\dfrac{4}{2}=2\] or \[t=\dfrac{-2}{2}=-1\]
Hence the roots of the equation are \[t=2\] or \[t=-1\]. But we substituted t for the term \[\sec (x)\].
\[\Rightarrow \sec (x)=2\] or \[\sec (x)=-1\]
We know that the general solution for \[x={{\sec }^{-1}}(a)\], here a is a real number in the range of \[\left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)\] is \[2n\pi \pm \theta \]. Where \[\theta \] is the solution in the principal range. The principal range for the inverse trigonometric function \[x={{\sec }^{-1}}(a)\] is \[\left[ 0,\dfrac{\pi }{2} \right)\bigcup \left( \dfrac{\pi }{2},\pi \right]\].
For \[\sec (x)=2\], the solution in the principal range is \[\theta =\dfrac{\pi }{3}\]. Hence, the general solution for \[\sec (x)=2\] is \[2n\pi \pm \dfrac{\pi }{3}\]. Here n is an integer.
For \[\sec (x)=-1\], the solution in the principal range is \[\theta =\pi \]. Hence, the general solution for \[\sec (x)=-1\] is \[2n\pi \pm \pi \]. Here n is an integer.

Note: To solve these types of problems, where we have a polynomial function in terms of another algebraic, trigonometric function, we should use the substitution method. It makes the equation easier to solve. The principal range of inverse trigonometric functions should be remembered, also their general solution.