Question

Write the number of the values of θ in $\left[ \theta ,2\pi \right]$ that satisfy the equation ${{\sin }^{2}}\theta -\cos \theta =\dfrac{1}{4}$.

Answer 1

Hint: In this given question, we have to express all trigonometric ratios in terms of $\cos \theta $. Thereafter, we can solve the resulting quadratic equation with the variable $\cos \theta $ to obtain the values of $\cos \theta $. As, $\cos \theta $ is an even function of θ, there will be two angles in $\left[ \theta ,2\pi \right]$ having the same cosine value. Using the obtained value of $\cos \theta $ we can get the required values of θ.

Complete step-by-step answer:

In this given question, we are asked to find the number of the values of θ in $\left[ \theta ,2\pi \right]$ that satisfy the equation ${{\sin }^{2}}\theta -\cos \theta =\dfrac{1}{4}............(1.1)$.

We know that,

${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$

$\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta ............(1.2)$

Now, substituting the value of ${{\sin }^{2}}\theta $ in equation 1.1 by $1-{{\cos }^{2}}\theta

$ as obtained from equation 1.2, we get

$\begin{align}

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

 & \Rightarrow 1-{{\cos }^{2}}\theta -\cos \theta =\dfrac{1}{4} \\

 & \Rightarrow {{\cos }^{2}}\theta +\cos \theta +\dfrac{1}{4}-1=0 \\

 & \Rightarrow {{\cos }^{2}}\theta +\cos \theta -\dfrac{3}{4}=0...........(1.3) \\

\end{align}$

We know that the quadratic formula for solution of x in the general quadratic equation $a{{x}^{2}}+bx+c=0$ is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}............(1.4)$.

Now, to get the value of $\cos \theta $, we solve equation 1.3 for $\cos \theta $ using the quadratic formula in equation 1.4.

$\begin{align}

  & {{\cos }^{2}}\theta +\cos \theta -\dfrac{3}{4}=0 \\

 & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 1\times \left( -\dfrac{3}{4} \right)}}{2\times 1} \\

 & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{4}}{2} \\

 & \Rightarrow \cos \theta =\dfrac{-1\pm 2}{2} \\

 & \Rightarrow \cos \theta =-\dfrac{3}{2}\text{ or }\dfrac{1}{2} \\

\end{align}$

So, we have got the values of $\cos \theta $ as $-\dfrac{3}{2}\text{ or }\dfrac{1}{2}$.

However, as the value of $\cos \theta $ must lie between -1 and 1, therefore the value of $\dfrac{-3}{2}$ should be discarded. Therefore, the value of $\cos \theta $ should be $\dfrac{1}{2}$.

Now, we know that as $\cos \theta $ is an even function, and its periodicity is $2\pi $, if an angle $\theta $ has a given cosine value, then the angle $2\pi -\theta $ will also have the same cosine value. Therefore, there should be two angles in $\left[ \theta ,2\pi \right]$ which would have the cosine value of $\dfrac{1}{2}$. Equation (1.3) and hence equation (1.1) will be satisfied for both these values of the angle $\theta $.

Therefore, the answer to the question is 2, i.e. there would be two values of $\theta $in $\left[ \theta ,2\pi \right]$ for which the equation given in the question will be satisfied.



Note: As we were asked to find only the number of values of $\theta $, and not the values of the angles, we did not explicitly obtain the value of $\theta $. However, it is important to obtain the value of $\cos \theta $ to check whether it is valid or not i.e. it lies between -1 and 1 or not. For each valid value of $\cos \theta $, there would be two angles in $\left[ \theta ,2\pi \right]$ which have the given $\cos \theta $ value.