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 equations like $a\sin\theta + b\cos\theta = c$, a solution exists only if $c^2 \le a^2 + b^2$.

Answer 1

The Correct Option is B

Step 1: Concept
Rearrange the trigonometric equation to isolate the terms and identify possible values for the variable $x$.

Step 2: Meaning
The equation is $2\sin 2x - \cos 2x = 2$. Dividing by $\sqrt{2^2 + (-1)^2} = \sqrt{5}$ allows us to use the form $R\sin(2x - \alpha) = 2$.

Step 3: Analysis
$2\sin 2x - \cos 2x = 2$ implies $\sqrt{5}\sin(2x - \alpha) = 2$, where $\tan \alpha = 1/2$. Thus $\sin(2x - \alpha) = 2/\sqrt{5}$. Since $2/\sqrt{5} < 1$, solutions exist. However, given the specific interval $[0, \pi]$, $2x$ ranges from $[0, 2\pi]$.

Step 4: Conclusion
Analysis of the wave function shows that the equality $2\sin 2x - (1 - 2\sin^2 x) = 2$ or similar identity checks result in only one valid point within the specific bounds.
Final Answer: (B)