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: Unit Circle Method:
Put $u=\cos x,\ v=\sin x$. We need $u-v=-1$ with $u^2+v^2=1$. Substituting $u=v-1$ gives $2v^2-2v=0$, so $v=0$ or $v=1$.

Step 2: Points:
$v=0$ gives $u=-1$, so $x=\pi+2n\pi$. $v=1$ gives $u=0$, so $x=\pi/2+2n\pi$.

Step 3: Filter:
In $[0,6\pi]$ the multiples of $\pi/3$ that appear are $\pi,3\pi,5\pi$, with $k=3,9,15$. The values $\pi/2+2n\pi$ give half-integer $k$. Option (A), 3 values.

Final Answer:
Option (A). \[ \boxed{\text{(A) } 3} \]