Question

Number of solutions of the equation f(x) = g(x) in x $\in$ [-π, $\frac{7\pi}{2}$] is equal to

• A. 4
• B. 5
• C. 6
• D. 8

Answer 1

Answer: (C)

6

Answer 2

The equation $f(x) = g(x)$ simplifies to $\text{cosx} + \text{sinx} - 1 = \text{sin}2\text{x} - 2$.
Substitute $t = \text{cosx} + \text{sinx}$, which implies $\text{sin}2\text{x} = t^2 - 1$.
The equation becomes $t - 1 = (t^2 - 1) - 2$, leading to $t^2 - t - 2 = 0$.
Solving for $t$, we get $(t-2)(t+1) = 0$, so $t=2$ or $t=-1$.
$t = \text{cosx} + \text{sinx} = \sqrt{2}\text{sin}(x + \frac{\pi}{4})$.
If $t=2$, $\sqrt{2}\text{sin}(x + \frac{\pi}{4}) = 2 \implies \text{sin}(x + \frac{\pi}{4}) = \sqrt{2}$, which has no solution as $\text{sin}(X) \le 1$.
If $t=-1$, $\sqrt{2}\text{sin}(x + \frac{\pi}{4}) = -1 \implies \text{sin}(x + \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.
This yields $x + \frac{\pi}{4} = 2n\pi + \frac{5\pi}{4}$ or $x + \frac{\pi}{4} = 2n\pi + \frac{7\pi}{4}$.
Solving for $x$, we get $x = (2n+1)\pi$ or $x = 2n\pi + \frac{3\pi}{2}$.
Listing solutions in $x \in [-\pi, \frac{7\pi}{2}]$:
For $x = (2n+1)\pi$: $-\pi, \pi, 3\pi$.
For $x = 2n\pi + \frac{3\pi}{2}$: $-\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}$.
Total distinct solutions: 6.