Question

The number of solutions in \([0,\frac{π}{2})\) of the equation \(cos3x\cdot tan5x = sin7x\) is

• A. \(4\)
• B. \(5\)
• C. \(7\)
• D. \(6\)

Hint:

Convert tan to sine over cosine, use product-to-sum formulas to get sin 8x = sin 12x, then count the solutions in the interval.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To solve a mixed trigonometric equation we write everything in sines and cosines and use product-to-sum formulas to reduce it to an equation of the form \(\sin A = \sin B\).

Step 2: Rewrite the equation.
\(\cos 3x \cdot \dfrac{\sin 5x}{\cos 5x} = \sin 7x\), with \(\cos 5x \neq 0\). Multiply by \(\cos 5x\):
\[ \cos 3x \sin 5x = \sin 7x \cos 5x \]

Step 3: Apply product to sum.
\[ \tfrac{1}{2}[\sin 8x + \sin 2x] = \tfrac{1}{2}[\sin 12x + \sin 2x] \]
So \(\sin 8x = \sin 12x\).

Step 4: Solve \(\sin 8x = \sin 12x\).
Either \(12x = 8x + 2k\pi\), giving \(x = k\pi/2\), or \(12x = \pi - 8x + 2k\pi\), giving \(x = (2k+1)\pi/20\).

Step 5: Count in \([0, \pi/2)\).
From the first family, only \(x = 0\) lies in the interval. From the second family, \((2k+1)\pi/20 < \pi/2\) means \(2k+1 < 10\), so \(2k+1 = 1, 3, 5, 7, 9\). This gives 5 values.

Step 6: Check \(\cos 5x \neq 0\).
\(\cos 5x = 0\) at \(x = \pi/10, 3\pi/10, \ldots\), which are \(2\pi/20, 6\pi/20, \ldots\), even multiples of \(\pi/20\). None of our 6 solutions is one of these, and \(x = 0\) works since both sides are 0.

Final Answer:
The total number of solutions is \(1 + 5 = 6\), option (D). \[ \boxed{6} \]