Question

How do you solve \[\sec x\sin x = 2\sin x\] from \[\left[ {0,2\pi } \right]\] ?

Answer 1

Hint: In order to solve this, first we will use the formula as \[\sec x = \dfrac{1}{{\cos x}}\] and rewrite the given equation. Then we will define the domain where this equation is defined. Since \[\cos x\] will be in the denominator, so our domain should exclude the points where \[\cos x = 0\] . After that we will divide by \[\sin x\] on both sides of the equation and try to find out the solutions of the given equation. But while dividing we will make sure that we don’t lose any solution, in particular those values of \[x\] where \[\sin x = 0\] .Hence, we will get the required solutions.

Complete step by step answer:
We have given,
\[\sec x\sin x = 2\sin x\]
And we have to find the solutions of the given equation.
Now we know that
\[\sec x = \dfrac{1}{{\cos x}}\]
Therefore, we get
\[\dfrac{{\sin x}}{{\cos x}} = 2\sin x{\text{ }} - - - \left( i \right)\]
Now, first of all we will define the domain where the above equation is defined. Since, \[\cos x\] is in the denominator, therefore our domain should exclude those points from the given interval i.e., \[\left[ {0,2\pi } \right]\] where \[\cos x = 0\].

So, in the interval \[\left[ {0,2\pi } \right]\]. \[\cos x = 0\] when \[x = \dfrac{\pi }{2}\] and \[x = \dfrac{{3\pi }}{2}\]. Therefore, we will restrict our solutions to these conditions.It means \[x \ne \dfrac{\pi }{2}\] and \[x \ne \dfrac{{3\pi }}{2}\]. Now, if we see from equation \[\left( i \right)\] both parts of the equation contain \[\sin x\]. So, we will divide by \[\sin x\] on both sides of the equation to simplify further. But while doing this, we have to make sure that we will not lose any solutions, particularly those values of \[x\] where \[\sin x = 0\]

As they are the solutions of the original equation since both parts of the equation would be equal to zero, but they might not be among those solutions of the new equation.So, from \[\sin x = 0\] we have three solutions as: \[x = 0,{\text{ }}x = \pi \] and \[x = 2\pi \] within the given interval \[\left[ {0,2\pi } \right]\]. Now on dividing by \[\sin x\] on both sides of equation \[\left( i \right)\] we get,
\[\dfrac{1}{{\cos x}} = 2\]
\[ \Rightarrow \cos x = \dfrac{1}{2}\]
which results in more two solutions as:
\[x = \dfrac{\pi }{3}\] and \[x = \dfrac{{5\pi }}{3}\] within the given interval \[\left[ {0,2\pi } \right]\]

Therefore, we get five solutions of the equation as \[x = 0,{\text{ }}x = \dfrac{\pi }{3},{\text{ }}x = \pi ,{\text{ }}x = \dfrac{{5\pi }}{3},{\text{ }}x = 2\pi \].

Note: Finding the domain of an equation is a tricky work that often leads to errors and mistakes. Therefore, this be carefully checked. Also do not forget to check the solutions from the original equation as they also play an important role in finding the solutions. Also, after finding the solutions, you can check whether the solutions obtained are correct or not by substituting them in the given equation.