Question

If the equation \(sin3θ-cos^2θ = \frac{1}{4}\) and \(θ\in [0,π]\), then the number of solutions is...

• A. \(0\)
• B. \(2\)
• C. \(4\)
• D. \(6\)

Hint:

Study \(f(\theta)=\sin3\theta-\cos^2\theta-\frac14\) on \([0,\pi]\) and count where it crosses zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
We need the number of roots of \(\sin3\theta=\cos^2\theta+\tfrac14\) in \([0,\pi]\). The right side lies between \(\tfrac14\) and \(\tfrac54\), so a root needs \(\sin3\theta\ge\tfrac14\).

Step 2: Key Formula or Approach
Compare the curves \(y=\sin3\theta\) and \(y=\cos^2\theta+\tfrac14\) on \([0,\pi]\). The sine curve has 3 half-waves, positive on \((0,\pi/3)\), negative on \((\pi/3,2\pi/3)\) and positive on \((2\pi/3,\pi)\).

Step 3: Detailed Explanation
On \((0,\pi/3)\): at \(\theta=0\), \(f=-1.25\). At \(\theta=\pi/6\), \(\sin3\theta=1\) and \(\cos^2\theta+\tfrac14=1\), so \(f=0\) exactly, and the maximum of \(f\) is only slightly above zero near here. The curve rises through zero and comes back through zero, giving 2 roots in this lobe.
On \((\pi/3,2\pi/3)\) the sine is negative, so there is no root.
On \((2\pi/3,\pi)\) the same pattern repeats, since \(\cos^2\theta\) is symmetric about \(\pi/2\) and \(\sin3\theta\) is symmetric under \(\theta\to\pi-\theta\) in magnitude. This gives 2 more roots.
Total roots: 4.

Final Answer:
The equation has 4 solutions in \([0,\pi]\), option (C). \[ \boxed{4\ \text{(C)}} \]