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,

\[(1 - \cos^2 x) - \cos x = \frac{1}{4}\]

\[ \Rightarrow 4 - 4 \cos^2 x - 4 \cos x = 1\]

\[ \Rightarrow 4 \cos^2 x + 4 \cos x - 3 = 0\]

\[ \Rightarrow 4 \cos^2 x + 6 \cos x - 2 \cos x - 3 = 0\]

\[ \Rightarrow 2 \cos x (2 \cos x + 3) - 1 (2 \cos x + 3) = 0\]

\[ \Rightarrow (2 \cos x + 3) ( 2 \cos x - 1) = 0\]
Here,
\[2 \cos x + 3 = 0\]

`=> cosx=-3/2` is not possible.
Or,
\[2 \cos x - 1 = 0\]
\[ \Rightarrow \cos x = \frac{1}{2}\]
\[ \Rightarrow \cos x = \cos \frac{\pi}{3}\]
\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}\]
Taking positive sign, 
\[x = \frac{7\pi}{3}, \frac{13\pi}{3}, \frac{19\pi}{3}, . . .\]
Taking negative sign, 

\[x = \frac{5\pi}{3}, \frac{11\pi}{3}, \frac{17\pi}{3}, . . .\]
`x=(5x)/3` and `(7x)/3`
will  satisfy the given condition, i.e., x in [0, 2π].
Hence, two values will satisfy the given equation.