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

Concept: High-power trigonometric equations can be solved by analyzing the natural bounds of the sine and cosine functions. For any real value of $x$, both $\sin^2 x$ and $\cos^2 x$ are constrained between 0 and 1. Raising a fraction less than 1 to a higher power shrinks its value ($y^p \le y^1$). Step 1: Set up bounding inequalities for the individual terms.
Using the power properties of numbers between 0 and 1: $$\sin^{100}x \le \sin^2 x$$ $$\cos^{100}x \ge 0 \quad \Rightarrow \quad -\cos^{100}x \le 0$$

Step 2: Combine the inequalities to evaluate the entire expression.
Summing our two bounding inequalities together: $$\sin^{100}x - \cos^{100}x \le \sin^2 x - 0 = \sin^2 x$$ We know from the Pythagorean identity that $\sin^2 x \le 1$. Therefore, the entire expression is strictly bounded by: $$\sin^{100}x - \cos^{100}x \le 1 \quad \cdots (1)$$

Step 3: Isolate the conditions required to reach equality.
The given problem states that the expression is exactly equal to its absolute upper limit of $1$: $$\sin^{100}x - \cos^{100}x = 1$$ Comparing this with equation (1), this scenario can only happen if both of our boundary conditions are met simultaneously:
• $\sin^{100}x = \sin^2 x = 1 \implies \sin x = \pm 1$
• $\cos^{100}x = 0 \implies \cos x = 0$

Step 4: Write out the general trigonometric solution.
On the unit circle, the conditions $\sin x = \pm 1$ and $\cos x = 0$ occur exclusively at the vertical positions: $$x = \pm\frac{\pi}{2}, \ \pm\frac{3\pi}{2}, \ \pm\frac{5\pi}{2}, \ \dots$$ We can write this complete sequence using standard general solution notation as: $$x = n\pi \pm \frac{\pi}{2}, \quad n \in \mathbb{I}$$ This matches option (B) perfectly.