Question

Solve the equation, \[{{\sin }^{2}}\theta -\cos \theta =\dfrac{1}{4}\].

Answer 1

Hint: At first try to change all the \[\sin \theta \] functions to \[\cos \theta \] function then use the completing the square method and find the values of \[\cos \theta +\dfrac{1}{2}\] value, then find \[\theta \].

Complete step-by-step answer:

In the question we are given an equation which is \[{{\sin }^{2}}\theta -\cos \theta =\dfrac{1}{4}\] and we have to find general solutions or values of \[\theta \].

So, we are given an equation,

\[{{\sin }^{2}}\theta -\cos \theta =\dfrac{1}{4}\]

At first we will change the \[\sin \theta \] to \[\cos \theta \] to maintain a single uniform trigonometric ratio to ease the problem.

We will use the identity that \[{{\sin }^{2}}\theta \] is equal to \[1-{{\cos }^{2}}\theta \] so on applying we get,

\[1-{{\cos }^{2}}\theta -\cos \theta =\dfrac{1}{4}\]

So, on rearranging the equation we get,

\[{{\cos }^{2}}\theta +\cos \theta =\dfrac{3}{4}\]

Now, we will add by \[\dfrac{1}{4}\] to both the sides so we get,

\[{{\cos }^{2}}\theta +\cos \theta +\dfrac{1}{4}=1\]

We can rewrite equation as,

\[{{\cos }^{2}}\theta +2\times \cos \theta \times \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}=1\]

We know a formula, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]. So, we can also use this as \[{{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}\], where, \[a=\cos \theta \] and \[b=\dfrac{1}{2}\]. Hence, we can write it as,

\[{{\left( \cos \theta +\dfrac{1}{2} \right)}^{2}}=1\]

So, we can write that \[\left( \cos \theta +\dfrac{1}{2} \right)\] is equal to \[\pm 1\].

At first take the care where the value of \[\left( \cos \theta +\dfrac{1}{2} \right)\] is 1 so we get,

\[\cos \theta +\dfrac{1}{2}=1\]

\[\cos \theta =\dfrac{1}{2}\]

We can write \[\dfrac{1}{2}\] as \[\cos \dfrac{\pi }{3}\]. So, \[\cos \theta =\cos \dfrac{\pi }{3}\].

Hence, the general solution is \[\theta =2n\pi \pm \dfrac{\pi }{3}\], where n is any integer using formula if \[\cos \theta =\cos \alpha \], then \[\theta =2n\pi \pm \alpha \]. Now let’s take the another case where value of \[\left( \cos \theta +\dfrac{1}{2} \right)\] is -1 so we get,

\[\cos \theta +\dfrac{1}{2}=-1\]

\[\cos \theta =\dfrac{-3}{2}\]

This case is not possible because \[\cos \theta \] can only attain values between -1 to 1 in which \[\dfrac{-3}{2}\] does not like.

Hence, the general solution is \[2n\pi \pm \dfrac{\pi }{3}\].

Note: Instead of doing like this one can also solve the equation by taking \[\cos \theta =t\] and then use the quadratic equation formula to find roots and hence substituting t as \[\cos \theta \] and find general solutions.