Question

Write the number of values of θ in [0, 2π] that satisfy the equation $\sin \mathrm{\theta }-\cos \mathrm{\theta }=\frac{1}{4}$.

Answer 1

Given equation: ${\sin }^2\theta -\cos \theta =\frac{1}{4}$
Now,
$$\begin{gathered} (1-{\cos }^2\theta )-\cos \theta =\frac{1}{4} \\ \Rightarrow 4-4{\cos }^2\theta -4\cos \theta =1 \\ \Rightarrow 4{\cos }^2\theta +4\cos \theta -3=0 \\ \Rightarrow 4{\cos }^2\theta +6\cos \theta -2\cos \theta -3=0 \\ \Rightarrow 2\cos \theta (2\cos \theta +3)-1(2\cos \theta +3)=0 \\ \Rightarrow (2\cos \theta +3)(2\cos \theta -1)=0 \end{gathered}$$
Here, $2\cos \theta +3=0$$\Rightarrow \cos \theta =-\frac{3}{2}$ is not possible.
Or,

$$\begin{gathered} 2\cos \theta -1=0 \\ \Rightarrow \cos \theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\pi \pm \frac{\mathrm{\pi }}{3} \end{gathered}$$

Taking positive sign, $\theta =\frac{7\mathrm{\pi }}{3},\frac{13\mathrm{\pi }}{3},\frac{19\mathrm{\pi }}{3},...$

Taking negative sign, $\theta =\frac{5\mathrm{\pi }}{3},\frac{11\mathrm{\pi }}{3},\frac{17\mathrm{\pi }}{3},...$

$\theta =\frac{5\mathrm{\pi }}{3}$ and $\frac{7\mathrm{\pi }}{3}$ will satisfy the given condition, i.e., $\theta$ in [0, 2π].

Hence, two values will satisfy the given equation.