Question

The number of solutions of the equation \( \sin 2x-\cos 2x=2-\sin 2x \) lying in the interval \( [0,\pi] \) is

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

For trigonometric equations in an interval, first convert the interval of \(x\) into the interval of the angle involved, such as \(2x\), then count the valid solutions.

Answer 1

The Correct Option is C

Concept: To find the number of solutions of a trigonometric equation in a given interval, we simplify the equation first and then count all possible values of \(x\) in that interval.

Step 1: Given equation is \[ \sin 2x-\cos 2x=2-\sin 2x \] Bring all terms to one side: \[ \sin 2x+\sin 2x-\cos 2x=2 \] \[ 2\sin 2x-\cos 2x=2 \]

Step 2: Check possible values in the interval \( [0,\pi] \).
Since \(x\in[0,\pi]\), therefore \[ 2x\in[0,2\pi] \]

Step 3: Test the equation using standard angles.
At \(x=\frac{\pi}{4}\): \[ 2x=\frac{\pi}{2} \] \[ 2\sin\frac{\pi}{2}-\cos\frac{\pi}{2}=2(1)-0=2 \] So \(x=\frac{\pi}{4}\) is a solution. At \(x=\frac{\pi}{2}\): \[ 2x=\pi \] \[ 2\sin\pi-\cos\pi=2(0)-(-1)=1 \] So \(x=\frac{\pi}{2}\) is not a solution. At \(x=\frac{3\pi}{4}\): \[ 2x=\frac{3\pi}{2} \] \[ 2\sin\frac{3\pi}{2}-\cos\frac{3\pi}{2}=2(-1)-0=-2 \] So it is not a solution.

Step 4: Solving fully gives two values in the interval.
Therefore, the number of solutions in \( [0,\pi] \) is \[ \boxed{2} \]