Trigonometric Equations — JEE Mains Mathematics

Formulas (2)

• Solve for the ratio, then count per period · General solutions
  $$\sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha;\qquad \cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha$$
• Two equations, domain limits and a sized interval · Roots of cos θ = c in order from 0
  $$\alpha,\ 2\pi-\alpha,\ 2\pi+\alpha,\ 4\pi-\alpha,\ \dots\qquad(\alpha=\cos^{-1}c\in(0,\pi))$$

Watch out for (2)

• Values ±1 and the endpoints
  → Solve for the ratio, then count per period
• Do not cancel a ratio
  → Two equations, domain limits and a sized interval

Formulas (2)

• Reduce to cos A = cos B · Two families
  $$\cos A=\cos B\Rightarrow A=2n\pi+B\ \text{or}\ A=2n\pi-B$$
• Collapse to one multiple angle · Triple angle
  $$\cos3x=4\cos^3x-3\cos x,\qquad \tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$$

Watch out for (2)

• Subtract the common angles
  → Reduce to cos A = cos B
• Check the original tangents
  → Collapse to one multiple angle

Formulas (2)

• The range decides whether a solution exists · Existence
  $$f(x)=k\ \text{has a solution exactly when}\ \min f\le k\le\max f$$
• a cos x + b sin x = c by the auxiliary angle · Auxiliary angle
  $$a\cos x+b\sin x=\sqrt{a^2+b^2}\,\cos(x-\varphi),\qquad\tan\varphi=\frac ba$$

Watch out for (2)

• The vertex may be outside
  → The range decides whether a solution exists
• Squaring adds roots
  → a cos x + b sin x = c by the auxiliary angle

Formulas (2)

• Powers with sin² and cos² in the exponent · Substitution
  $$t=a^{\sin^2x}:\qquad a^{\sin^2x}+a^{\cos^2x}=t+\frac at$$
• Bounds, graphs and moduli · Forced equality
  $$f(x)\le m\le g(x)\ \text{for all}\ x:\quad f(x)=g(x)\ \text{exactly when}\ f(x)=g(x)=m$$

Watch out for (2)

• A root in t may be out of range
  → Powers with sin² and cos² in the exponent
• Check the part-branches at the ends
  → Bounds, graphs and moduli