Trigonometric Equations — Counting Solutions
DAILY_CHALLENGE
Grade None
Question:
Match each entry in List-I to the correct entry in List-II and choose the correct option.
**List-I**
(P) The number of elements in $\{x\in[-\pi,\pi]:\ \sin^6 x+\cos^4 x=1\}$
(Q) The number of elements in $\left\{x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]:\ \sin^2 x+\cos^6 x=1\right\}$
(R) The number of elements in $\left\{x\in[-\pi,\pi]:\ \cos^2\!\dfrac{x}{2}-\sin^2 x=\dfrac{1}{2}\right\}$
(S) The number of elements in $\left\{x\in[-2\pi,2\pi]:\ 6\sin^2\!\dfrac{x}{2}-\cos 3x=3\right\}$
**List-II**
(1) is 1
(2) is 2
(3) is 3
(4) is 4
(5) is 5
$(P)\to(2),\ (Q)\to(5),\ (R)\to(3),\ (S)\to(4)$
$(P)\to(5),\ (Q)\to(3),\ (R)\to(2),\ (S)\to(4)$
$(P)\to(5),\ (Q)\to(4),\ (R)\to(1),\ (S)\to(3)$
$(P)\to(4),\ (Q)\to(3),\ (R)\to(2),\ (S)\to(5)$
Step-by-Step Solution
Key Concept: For equations of the form $\text{trig}^m + \text{trig}^n = 1$ with $m > n$, factor using $a^m - a^n = a^n(a^{m-n}-1)$ and the Pythagorean identity to split into two clean cases.
**Step 1: Solve P: $\sin^6x+\cos^4x=1$ on $[-\pi,\pi]$**
Rewrite: $\sin^6x = \sin^2x(1+\cos^2x)$, so $\sin^2x[\sin^4x-1-\cos^2x]=0$. Case 1: $\sin x=0 \Rightarrow x\in\{-\pi,0,\pi\}$ — 3 points. Case 2: $\sin^4x=1+\cos^2x$, forcing $\cos x=0 \Rightarrow x=\pm\pi/2$ — 2 points. Total: $5$. $P\to(5)$.
**Step 2: Solve Q: $\sin^2x+\cos^6x=1$ on $[-\pi/2,\pi/2]$**
$\sin^2x=\sin^2x(1+\cos^2x+\cos^4x)$. Case 1: $\sin x=0 \Rightarrow x=0$ — 1 point. Case 2: $\cos^2x(1+\cos^2x)=0 \Rightarrow \cos x=0 \Rightarrow x=\pm\pi/2$ — 2 points. Total: $3$. $Q\to(3)$.
**Step 3: Solve R: $\cos^2(x/2)-\sin^2x=1/2$ on $[-\pi,\pi]$**
Use $\cos^2(x/2)=\tfrac{1+\cos x}{2}$: equation becomes $2\cos^2x+\cos x-2=0$. Discriminant $= 1+16=17$. Valid root: $\cos x=\tfrac{-1+\sqrt{17}}{4}\approx0.78$, giving $x=\pm\arccos(0.78)$. Total: $2$. $R\to(2)$.
**Step 4: Solve S: $6\sin^2(x/2)-\cos3x=3$ on $[-2\pi,2\pi]$**
$3(1-\cos x)-\cos3x=3 \Rightarrow -3\cos x-\cos3x=0$. Using $\cos3x=4\cos^3x-3\cos x$: $-4\cos^3x=0 \Rightarrow \cos x=0$. Solutions: $x\in\{\pm\pi/2,\pm3\pi/2\}$ — 4 points. $S\to(4)$.
Correct Answer: B