Question

How do you solve the trigonometric equation \[2{\left( {\sin x} \right)^2} - \cos x - 1 = 0\] to find the values of x in the interval \[\left[ {0,2\pi } \right]\] that satisfy the given equation.

Answer 1

Hint: First convert \[{\sin ^2}x\] into \[{\cos ^2}x\] by the use of trigonometric identity then assume \[\cos x\] as \[y\] and obtain the quadratic equation. Solve the quadratic equation by the use of middle term splitting method and thus get the values of y. After that replace y as \[\cos x\] and then get the values of x.

Complete step by step solution:
We are given that we are required to solve \[2{\left( {\sin x} \right)^2} - \cos x - 1 = 0\].
So, first convert the sine term into cosine by the use of trigonometric formula \[{\sin ^2}x = 1 - {\cos ^2}x\] as shown below.
\[2\left\{ {1 - {{\left( {\cos x} \right)}^2}} \right\} - \cos x - 1 = 0\]
Simplify the equation as shown below.
\[ \Rightarrow 2 - 2{\left( {\cos x} \right)^2} - \cos x - 1 = 0\]
\[ \Rightarrow - 2{\left( {\cos x} \right)^2} - \cos x + 1 = 0\]
Multiply both sides of the equation by negative one and write the result.
\[ \Rightarrow 2{\left( {\cos x} \right)^2} + \cos x - 1 = 0\]
Let us assume that y = cos x.
Replacing this in the given equation, we will obtain the following equation: -
\[ \Rightarrow 2{y^2} + y - 1 = 0\]
Now, we can write the equation in the above line as following also: -
\[ \Rightarrow 2{y^2} + 2y - y - 1 = 0\]
Now, taking 2y common from the first two terms in the left hand side of above expression, we will then obtain the following expression: -
\[ \Rightarrow 2y\left( {y + 1} \right) - y - 1 = 0\]
Now, taking \[ - 1\] common from the last two terms in the left hand side of the above expression, we will then obtain the following expression: -
\[ \Rightarrow 2y\left( {y + 1} \right) - 1\left( {y + 1} \right) = 0\]
Now taking the factor (y+1) common from the above equation, we will then obtain the following expression: -
\[ \Rightarrow \left( {y + 1} \right)\left( {2y - 1} \right) = 0\]
Now, we have got two values of y which are \[\dfrac{1}{2}\] and \[ - 1\].
Putting back y as cos x as we assumed in the beginning, we will then obtain that the two possible values of cos (x) are \[\dfrac{1}{2}\] and \[ - 1\]. But since we know that sine of any function always lies between – 1 and 1 also the values of x in the interval \[\left[ {0,2\pi } \right]\] that satisfy the given equation.
Therefore, values of cos(x) that satisfy the equation \[\cos \left( x \right) = \dfrac{1}{2}\] and in the interval \[\left[ {0,2\pi } \right]\] are \[\dfrac{\pi }{6}\] and \[\dfrac{{5\pi }}{3}\].
Similarly, values of cos(x) that satisfy the equation \[\cos \left( x \right) = - 1\] and in the interval \[\left[ {0,2\pi } \right]\] is \[\pi \] only.
Thus with in the interval \[\left[ {0,2\pi } \right]\], the solutions for x are \[\dfrac{\pi }{6}\], \[\pi \] and \[\dfrac{{5\pi }}{3}\].

Note:
The students must note that in the given solution we used splitting of middle term method but you may also use the fact that the roots of \[a{y^2} + by + c = 0\] are given by:-
\[ \Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]