Question

The general solution of the equation \(\sin^{100}x-\cos^{100}x=1\) is:

• A. $\{2n\pi+\frac{\pi}{3}:n\in I\}$
• B. $\{n\pi\pm\frac{\pi}{2}:n\in I\}$
• C. $\{n\pi+\frac{\pi}{4}:n\in I\}$
• D. $\{2m\pi-\frac{\pi}{3}:n\in I\}$

Hint:

Whenever you see a high even power equation equal to 1, like $\sin^{2m}x - \cos^{2k}x = 1$, the negative term must vanish completely ($0$) and the positive term must reach its absolute peak ($1$). This allows you to instantly skip complex algebraic factoring and go straight to the root angles!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are dealing with high even powers of trigonometric functions. Note that for any real \( x \), \( \sin^2 x \le 1 \) and \( \cos^2 x \le 1 \). Consequently, raising these values to high powers like 100 generally makes them much smaller, except when they are exactly 0 or \( \pm 1 \).
Step 2: Key Formula or Approach:
1. Analyze the equation as \( \sin^{100} x = 1 + \cos^{100} x \).
2. Use the fact that \( \sin^{100} x \le \sin^2 x \le 1 \).
3. Compare the maximum possible value of the LHS with the minimum possible value of the RHS.
Step 3: Detailed Explanation:
LHS: \( \sin^{100} x \). Since \( |\sin x| \le 1 \), \( \sin^{100} x \le 1 \).
RHS: \( 1 + \cos^{100} x \). Since \( \cos^{100} x \ge 0 \), \( 1 + \cos^{100} x \ge 1 \).
For the equation to hold, the LHS must reach its maximum and the RHS must reach its minimum simultaneously: \[ \sin^{100} x = 1 \quad \text{and} \quad \cos^{100} x = 0 \] This occurs only when: \[ \sin^2 x = 1 \quad \text{and} \quad \cos^2 x = 0 \] Which implies: \[ \cos x = 0 \implies x = (2n + 1)\frac{\pi}{2} = n\pi \pm \frac{\pi}{2} \] At these points, \( \sin x = \pm 1 \), so \( \sin^{100} x = 1 \), satisfying the original equation.
Step 4: Final Answer:
The general solution is \( x = n\pi \pm \frac{\pi}{2} \).