If 4sinθ + cscθ = 4, 0° < θ < 90°, then the value of sin 3θ + cos 3θ is

Question

If 4sinθ + cscθ = 4, 0° < θ < 90°, then the value of sin 3θ + cos 3θ is

1. A. -1
2. B. 0
3. C. 1 (Correct answer)
4. D. 2

Correct Answer

Option C — 1

Detailed Solution & Explanation

The correct answer is 1.

Shortcut Trick

• The given equation is a perfect square in disguise. Writing $s = \sin\theta$:
  • $4s + \dfrac{1}{s} = 4 \Rightarrow 4s^2 - 4s + 1 = 0 \Rightarrow (2s - 1)^2 = 0$
• So $\sin\theta = \dfrac{1}{2}$, and since $0° < \theta < 90°$, $\theta = 30°$.
• Then $3\theta = 90°$, giving $\sin 90° + \cos 90° = 1 + 0 = \textbf{1}$.

Alternate Method

• Multiply through by $\sin\theta$: $4\sin^2\theta - 4\sin\theta + 1 = 0$.
• By the quadratic formula, the discriminant is $16 - 16 = 0$, so there is a repeated root $\sin\theta = \dfrac{4}{8} = \dfrac{1}{2}$.
• $\theta = 30°$ is the only value in the given range, and $\sin 3\theta + \cos 3\theta = \sin 90° + \cos 90° = \textbf{1}$.

Additional Information

• Note the expression is $\sin 3\theta + \cos 3\theta$ — three times the angle, not the cube of the sine. Reading it as $\sin^3\theta + \cos^3\theta$ gives $\dfrac{1 + 3\sqrt{3}}{8} \approx 0.77$, which is not among the options.
• $\csc\theta = \dfrac{1}{\sin\theta}$, so any equation mixing $\sin$ and $\csc$ becomes a quadratic once multiplied through.
• A zero discriminant is a strong hint the question was constructed around a standard angle — worth checking for whenever the algebra collapses this neatly.