Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\left(2 \sin x - \cos x\right) \left(1 + \cos x\right) = \sin^2 x$

Given equation: $\;\;\;$ $\left(2 \sin x - \cos x\right) \left(1 + \cos x\right) = \sin^2 x$

i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x - \cos^2 x = \sin^2 x$

i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x = \sin^2 x + \cos^2 x$

i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x = 1$

i.e. $\;$ $2 \sin x \left(1 + \cos x\right) - \left(1 + \cos x\right) = 0$

i.e. $\;$ $\left(2 \sin x - 1\right) \left(1 + \cos x\right) = 0$

i.e. $\;$ $2 \sin x - 1 = 0$ $\;\;\;$ or $\;\;\;$ $1 + \cos x = 0$

When $\;\;$ $2 \sin x - 1 = 0$

i.e. $\;$ $\sin x = \dfrac{1}{2}$

$\implies$ $x = \left(-1\right)^n \sin^{-1} \left(\dfrac{1}{2}\right) + n \pi, \;\;\; n \in Z$

i.e. $\;$ $x = \left(-1\right)^n \times \dfrac{\pi}{6} + n \pi, \;\;\; n \in Z$

Or when $\;\;$ $1 + \cos x = 0$

i.e. $\;$ $\cos x = -1$

$\implies$ $x = \pm \cos^{-1} \left(-1\right) + 2 m \pi, \;\;\; m \in Z$

i.e. $\;$ $x = 2 m \pi \pm \pi, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $2 \sin x + \tan x = 0$

Given equation: $\;\;\;$ $2 \sin x + \tan x = 0$

i.e. $\;$ $2 \sin x + \dfrac{\sin x}{\cos x} = 0$

i.e. $\;$ $2 \sin x \cos x + \sin x = 0$

i.e. $\;$ $\sin x \left(2 \cos x + 1\right) = 0$

$\implies$ $\sin x = 0$ $\;\;\;$ or $\;\;\;$ $2 \cos x + 1 = 0$

When $\;\;$ $\sin x = 0$

$\implies$ $x = n \pi, \;\;\; n \in I$

When $\;\;$ $2 \cos x + 1 = 0$

$\implies$ $\cos x = \dfrac{-1}{2} = \cos \left(\pi - \dfrac{\pi}{3}\right) = \cos \left(\dfrac{2 \pi}{3}\right)$

i.e. $\;$ $x = 2 m \pi \pm \dfrac{2 \pi}{3}, \;\;\; m \in I$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\sqrt{2} \cos^2 \left(7x\right) - \cos \left(7x\right) = 0$

Given equation: $\;\;\;$ $\sqrt{2} \cos^2 \left(7x\right) - \cos \left(7x\right) = 0$

i.e. $\;$ $\cos \left(7x\right) \left[\sqrt{2} \cos \left(7x\right) - 1\right] = 0$

$\implies$ $\cos \left(7x\right) = 0$ $\;\;$ or $\;\;$ $\sqrt{2} \cos \left(7x\right) - 1 = 0$

When $\;\;$ $\cos \left(7x\right) = 0$

$\implies$ $7x = \dfrac{\left(2n + 1\right) \pi}{2}, \;\;\; n \in I$

i.e. $\;$ $x = \dfrac{\pi}{14} \left(2n + 1\right), \;\;\; n \in I$

When $\;\;$ $\sqrt{2} \cos \left(7x\right) - 1 = 0$

i.e. $\;$ $\cos \left(7x\right) = \dfrac{1}{\sqrt{2}}$

i.e. $\;$ $7x = \pm \cos^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + 2 m \pi, \;\;\; m \in I$

i.e. $\;$ $7x = \pm \dfrac{\pi}{4} + 2 m \pi, \;\;\; m \in I$

i.e. $\;$ $x = \pm \dfrac{\pi}{28} + \dfrac{2}{7} m \pi, \;\;\; m \in I$

i.e. $\;$ $x = \dfrac{\pi}{28} \left(8m \pm 1\right), \;\;\; m \in I$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\sin \left(3x - 2\right) = -1$

Given equation: $\;\;\;$ $\sin \left(3x - 2\right) = -1$

i.e. $\;$ $3x - 2 = \sin^{-1} \left(-1\right)$

i.e. $\;$ $3x - 2 = - \sin^{-1} \left(1\right)$ $\;\;\;$ $\left\{\because \;\; \sin^{-1} \left(-x\right) = - \sin^{-1} x \;\; \forall \; x \in \left[-1, 1\right]\right\}$

i.e. $\;$ $3x - 2 = \dfrac{- \pi}{2} + 2 n \pi \;\;\; n \in Z$

i.e. $\;$ $3x = \dfrac{4 n \pi - \pi}{2} + 2 \;\;\; n \in Z$

i.e. $\;$ $x = \dfrac{4n \pi - \pi}{6} + \dfrac{2}{3} \;\;\; n \in Z$

i.e. $\;$ $x = \dfrac{\pi \left(4n - 1\right) + 4}{6} \;\;\; n \in Z$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{4}{5}\right) - \cos^{-1} \left(\dfrac{2}{\sqrt{5}}\right) = \tan^{-1} \left(\dfrac{1}{2}\right)$

Formulas

$\sin^{-1} x = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \;\;\; -1 < x < 1$

$\cos^{-1} x = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right), \;\;\; -1 < x < 0, \; 0 < x < 1$

$\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right), \;\;\; x \cdot y < 1$

$\begin{aligned} LHS & = \sin^{-1} \left(\dfrac{4}{5}\right) - \cos^{-1} \left(\dfrac{2}{\sqrt{5}}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{4}{5}}{\sqrt{1 - \dfrac{16}{25}}}\right) - \tan^{-1} \left(\dfrac{\sqrt{1 - \dfrac{4}{5}}}{\dfrac{2}{\sqrt{5}}}\right) \\\\ & = \tan^{-1} \left(\dfrac{4}{3}\right) - \tan^{-1} \left(\dfrac{1}{2}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{4}{3} - \dfrac{1}{2}}{1 + \dfrac{4}{3} \times \dfrac{1}{2}}\right) \\\\ & = \tan^{-1} \left(\dfrac{5}{10}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{2}\right) = RHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) = \dfrac{\pi}{2}$

$\begin{aligned} LHS & = \sin^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{4}{5} \times \sqrt{1 - \dfrac{25}{169}} + \dfrac{5}{13} \times \sqrt{1 - \dfrac{16}{25}}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\ & \left\{\because \;\; \sin^{-1} x + \sin^{-1} y = \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right), \; 0 < x, y < 1\right\} \\\\ & = \sin^{-1} \left(\dfrac{4}{5} \times \dfrac{12}{13} + \dfrac{5}{13} \times \dfrac{3}{5}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65} \times \sqrt{1 - \dfrac{256}{4225}} + \dfrac{16}{65} \times \sqrt{1 - \dfrac{3969}{4225}}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65} \times \sqrt{\dfrac{3969}{4225}} + \dfrac{16}{65} \times \sqrt{\dfrac{256}{4225}}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65} \times \dfrac{63}{65} + \dfrac{16}{65} \times \dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{4225}{4225}\right) \\\\ & = \sin^{-1} \left(1\right) \\\\ & = \dfrac{\pi}{2} = RHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\tan^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) = \tan^{-1} \left(3 + 2 \sqrt{2}\right)$

$\begin{aligned} LHS & = \tan^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{1 - \dfrac{1}{2}}}\right) \\ & \left\{\because \;\; \sin^{-1} x = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \; 0 < x < 1\right\} \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \tan^{-1} \left(1\right) \\\\ & = \tan^{-1} \left[\dfrac{\dfrac{1}{\sqrt{2}} + 1}{1 - \dfrac{1}{\sqrt{2}} \times 1}\right] \\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \; x \cdot y < 1\right\} \\\\ & = \tan^{-1} \left[\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}\right] \\\\ & = \tan^{-1} \left[\dfrac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1 \right) \left(\sqrt{2} + 1\right)}\right] \\\\ & = \tan^{-1} \left[\dfrac{2 + 1 + 2 \sqrt{2}}{2 - 1}\right] \\\\ & = \tan^{-1} \left[3 + 2 \sqrt{2}\right] = RHS \end{aligned}$