General Solutions & Counting Solutions

Write each equation in the form (function of θ) = (same function of a known angle α), then:

• $\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n \alpha$
• $\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$
• $\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$

for $n \in \mathbb{Z}$. The principal solution is the one in the first cycle; the general solution adds the periodic family. (Special: $\sin\theta=0 \Rightarrow \theta=n\pi$; $\cos\theta=0 \Rightarrow \theta=(2n+1)\tfrac{\pi}{2}$.) The identity layer underneath — how an equation gets INTO a solvable shape. Most PYQs here open with one of these before any general solution is written (full treatment in the Trigonometric Identities notes, linked below):

• Reciprocals: $\sec\theta=\tfrac1{\cos\theta}$, $\csc\theta=\tfrac1{\sin\theta}$, $\cot\theta=\tfrac1{\tan\theta}=\tfrac{\cos\theta}{\sin\theta}$. Pythagorean: $\sin^2\theta+\cos^2\theta=1$, $1+\tan^2\theta=\sec^2\theta$, $1+\cot^2\theta=\csc^2\theta$ — the tool for turning a mixed $\sin^2,\cos$ equation into a quadratic in ONE ratio.
• Double angle: $\sin2\theta=2\sin\theta\cos\theta$, $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$; power reduction $\sin^2\theta=\tfrac{1-\cos2\theta}{2}$, $\cos^2\theta=\tfrac{1+\cos2\theta}{2}$ (so $8\sin^2(\theta+\tfrac\pi4)=4(1+\sin2\theta)$).
• Compound angle: $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$, $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$, $\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ — the DIFFERENCE forms included. $\cos A\cos B-\sin A\sin B=\cos(A+B)$ is what collapses $\cot2x\cot3x=1$ to $\cos5x=0$.
• Co-functions: $\cos\theta=\sin(\tfrac\pi2-\theta)$ AND $\tan A=\cot B\Rightarrow A+B=\tfrac\pi2$ (mod $\pi$). Angle sum of a triangle $=\pi$, so "angles $x,x,x$" means $3x=\pi$.
• Half angle: $1+\cos x=2\cos^2\tfrac x2$, $1-\cos x=2\sin^2\tfrac x2$, $\sin x=2\sin\tfrac x2\cos\tfrac x2$; Weierstrass with $t=\tan\tfrac\theta2$: $\sin\theta=\dfrac{2t}{1+t^2}$, $\cos\theta=\dfrac{1-t^2}{1+t^2}$, $\tan\theta=\dfrac{2t}{1-t^2}$ — note the HALF angle; $t=\tan\theta$ is the recorded wrong substitution.

Common reductions:

• Co-function: $\sin 2\theta = \cos 3\theta = \sin(\tfrac{\pi}{2} - 3\theta)$, then equate angles.
• Quadratic in one ratio: equations like $2\cos^2 x + \cos x - 1 = 0$ factor; solve each linear factor.
• Half-angle collapse: $\csc x + \cot x = \dfrac{1+\cos x}{\sin x}$, and with $1+\cos x = 2\cos^2\tfrac{x}{2}$, $\sin x = 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ this is $\cot\tfrac{x}{2}$. So $\csc x + \cot x = \sqrt3$ is just $\cot\tfrac{x}{2} = \sqrt3\Rightarrow \tfrac{x}{2} = \tfrac{\pi}{6}\Rightarrow x = \tfrac{\pi}{3}$ on $(0,2\pi)$ — no squaring, no false roots. Its sibling: $\csc x - \cot x = \tan\tfrac{x}{2}$. (Cross-multiplying to $1+\cos x = \sqrt3\sin x$ and squaring also works, but manufactures $\cos x = -1$, which must then be discarded.)
• Weierstrass $t=\tan\tfrac x2$ when an equation mixes $\sin x$ and $\cos x$ linearly, e.g. $a\sin x+b\cos x=c$: $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ turn it into the quadratic $(b+c)t^2-2at+(c-b)=0$; each real $t$ gives $x=2\tan^{-1}t$ (and check $x=\pi$, where $t$ is undefined, separately).
• Always verify roots in the ORIGINAL equation — squaring or dividing can introduce or drop solutions (e.g. where a denominator vanishes).

To count solutions of (function) $= k$ on an interval:

• Reduce to one trig function equal to a constant (e.g. $\cot 2x\cot 3x = 1 \Rightarrow \cos 5x = 0$).
• Find the general solution, then list the values of $n$ that keep $\theta$ inside the interval.
• Discard values where the original equation is undefined (a cotangent/cosecant blowing up, a denominator zero).
• Graphically, the count is the number of intersections of $y = (\text{function})$ with the horizontal line $y = k$ over the interval.

• $\sin x, \cos x \in [-1, 1]$, so $a\sin x = b$ needs $|b| \le |a|$; count integer parameters by intersecting with this range.
• For $\cos^{100}x - \sin^{100}x = 1$: since both terms are in $[0,1]$, equality forces $\cos^2 x = 1$, i.e. $x = n\pi$.
• A boundary condition ($0 \le x \le \tfrac{\pi}{2}$) plus an undefined-point check often cuts the candidate solutions down to one or two.