Question

Let \(x\in [0,6π]\) satisfy the equation \(cosx-sinx = -1\). If \(x = k(\frac{π}{3})\) where \(k\in N\), then find the number of possible values of \(k\) is .....

• A. \(3\)
• B. \(4\)
• C. \(6\)
• D. \(8\)

Hint:

Solve for x in [0, 6 pi], then keep only the values that are whole multiples of pi/3.

Answer 1

The Correct Option is A

Step 1: Reduce the Equation:
Write \(\cos x-\sin x=\sqrt2\cos\left(x+\dfrac{\pi}{4}\right)\). So
\[ \cos\left(x+\frac{\pi}{4}\right)=-\frac{1}{\sqrt2} \]

Step 2: General Solution:
\(x+\dfrac{\pi}{4}=2n\pi\pm\dfrac{3\pi}{4}\). This gives \(x=2n\pi+\dfrac{\pi}{2}\) or \(x=2n\pi+\pi\).

Step 3: List Values in [0, 6 pi]:
From \(x=2n\pi+\pi/2\): \(\tfrac{\pi}{2},\ \tfrac{5\pi}{2},\ \tfrac{9\pi}{2}\).
From \(x=2n\pi+\pi\): \(\pi,\ 3\pi,\ 5\pi\).

Step 4: Keep Those of the Form k(pi/3):
Write each as \(k\pi/3\): \(\pi/2\) gives \(k=3/2\), \(5\pi/2\) gives \(15/2\), \(9\pi/2\) gives \(27/2\). None is a natural number.
\(\pi\) gives \(k=3\), \(3\pi\) gives \(k=9\), \(5\pi\) gives \(k=15\).
So \(k\) takes 3 values: 3, 9, 15.

Final Answer:
There are 3 possible values of \(k\), option (A). \[ \boxed{\text{(A) } 3} \]