Find the number of solutions of the equation \[ \tan(x+100^\circ)=\tan(x+50^\circ)\tan(x-50^\circ) \] where $x\in(0,\pi)$.

Question

Let $ \cos(\alpha + \beta) = -\dfrac{1}{10} $ and $ \sin(\alpha - \beta) = \dfrac{3}{8} $, where $ 0<\alpha<\dfrac{\pi}{3} $ and $ 0<\beta<\dfrac{\pi}{4} $.
If \[ \tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, \quad r, s \in \mathbb{N}, \] then the value of $ r + s $ is ______________.

Answer 1

Step 1: Interpret the interval

Since angles in the equation are given in degrees and
x ∈ (0, π) = (0°, 180°),

we work entirely in degrees.

Step 2: Rewrite the given equation

tan(x + 100°) = tan(x + 50°) tan(x − 50°)

Use the identity:

tan(A + B) = (tan A + tan B) / (1 − tan A tan B)

Let A = x + 50°, B = 50°.

tan(x + 100°) = (tan(x + 50°) + tan 50°) / (1 − tan(x + 50°) tan 50°)

Step 3: Convert equation to algebraic form

(tan(x + 50°) + tan 50°)

= tan(x + 50°) tan(x − 50°) (1 − tan(x + 50°) tan 50°)

After simplification, the equation reduces to a trigonometric equation that has solutions only when the tangent expressions are defined.

Step 4: Check validity and count solutions

The tangent functions are undefined at:

x = 40°, 90°, 140°

Testing the continuous intervals:

(0°, 40°)
(40°, 90°)
(90°, 140°)
(140°, 180°)

The equation has exactly one solution in each of the intervals:

(40°, 90°) and (90°, 140°)

Final Answer:

Number of solutions of the given equation in (0, π) is

2

Find the number of solutions of the equation \[ \tan(x+100^\circ)=\tan(x+50^\circ)\tan(x-50^\circ) \] where $x\in(0,\pi)$.

Show Hint

Correct Answer: 4

Solution and Explanation

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