Question

The general solution of $2\cos\theta - \sqrt{3} = 0$ is}

• A. $\theta = n\pi + (-1)^n \frac{\pi}{6}$
• B. $\theta = 2n\pi \pm \frac{\pi}{6}$
• C. $\theta = 2n\pi \pm \frac{\pi}{3}$
• D. $\theta = n\pi + (-1)^n \frac{\pi}{3}$

Hint:

Remember:
• $\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$
• $\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha$

Answer 1

The Correct Option is B

Concept: Trigonometric equations involving cosine functions have periodic solutions due to the cyclic nature of the cosine graph. The general solution depends on the principal value and the periodicity of the cosine function, which repeats every $2\pi$.

Step 1: Simplifying the given equation
We are given: \[ 2\cos\theta - \sqrt{3} = 0 \] First, isolate the trigonometric term: \[ 2\cos\theta = \sqrt{3} \] Now divide both sides by 2: \[ \cos\theta = \frac{\sqrt{3}}{2} \]

Step 2: Finding principal value
We know from standard trigonometric values that: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] Thus, the principal solution angle is: \[ \theta = \frac{\pi}{6} \]

Step 3: Applying general solution of cosine
For cosine equations, the general solution formula is: \[ \theta = 2n\pi \pm \alpha \] where $\alpha$ is the principal angle. Substituting $\alpha = \frac{\pi}{6}$: \[ \theta = 2n\pi \pm \frac{\pi}{6} \]

Step 4: Final interpretation
This represents all angles where cosine equals $\frac{\sqrt{3}}{2}$, including both symmetric positions in the unit circle. Thus, the correct answer is option (B).