Question

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

Answer 1

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

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

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

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

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

Hence, two values will satisfy the given equation.