Question

If \(5\pi - 6 \cos^{-1}(\sqrt{3}(2x - 1)) = 0\), then the value of \(x\) is equal to

• A. \(\frac{1}{2}\)
• B. 2
• C. \(\frac{1}{4}\)
• D. 4
• E. 0

Hint:

Remember the range of \(\cos^{-1}(x)\) is \([0, \pi]\). Since \(\frac{5\pi}{6}\) lies in this range, the equation is valid. Always look for potential sign errors when evaluating trig values in the 2nd quadrant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We isolate the inverse trigonometric term and then apply the trigonometric function to both sides to solve for \(x\).

Step 2: Key Formula or Approach:
If \(\cos^{-1}(\alpha) = \theta\), then \(\alpha = \cos \theta\).

Step 3: Detailed Explanation:
The equation is:
\[ 5\pi = 6 \cos^{-1}(\sqrt{3}(2x - 1)) \]
Divide by 6:
\[ \cos^{-1}(\sqrt{3}(2x - 1)) = \frac{5\pi}{6} \]
Take cosine on both sides:
\[ \sqrt{3}(2x - 1) = \cos(\frac{5\pi}{6}) \]
Since \(\frac{5\pi}{6} = 150^\circ\), which is in the second quadrant:
\[ \cos(\frac{5\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} \]
Now, equate the expressions:
\[ \sqrt{3}(2x - 1) = -\frac{\sqrt{3}}{2} \]
Divide by \(\sqrt{3}\):
\[ 2x - 1 = -\frac{1}{2} \]
\[ 2x = 1 - \frac{1}{2} = \frac{1}{2} \]
\[ x = \frac{1}{4} \]

Step 4: Final Answer:
The value of \(x\) is \(\frac{1}{4}\).