Trigonometric Equations

Trigonometry: (lesson 2 of 3)

Very, very important formula:

\begin{aligned} &\color{red}{1}:\ \sin \alpha = \sin \beta \\ &\color{red}{2}:\ \cos \alpha = \cos \beta \Rightarrow \alpha = \beta + 2k\pi \ or \\ &\hspace{4.5cm} \alpha = - \beta + 2k\pi \\ &\color{red}{3}:\ \tan \alpha = \tan \beta \Rightarrow \alpha = \beta + k\pi \end{aligned}

Example 1:

\begin{aligned} \sin (x) = \frac{1}{2} \end{aligned}

Solution: We know that the $\sin\frac{\pi}{6}=\frac{1}{2}$ and therefore

\begin{aligned} \sin (x) = \sin ( \frac{\pi }{6} )\mathop \to \limits^{\color{red}{formula1}} &x = \frac{\pi }{6} + 2k\pi \ or \\ &x = \frac{5\pi}{6} + 2k\pi \end{aligned}

Example 2:

\begin{aligned} \cos (x) = \frac{\sqrt 2 }{2} \end{aligned}

Solution: We know that the $\cos ( \frac{\pi}{4} ) = \frac{\sqrt 2}{2}$ and therefore

\begin{aligned} \cos (x) = \cos ( \frac{\pi }{4} )\mathop \to \limits^{\color{red}{formula2}} &x = \frac{\pi}{4} + 2k\pi \ or \\ &x = -\frac{\pi}{4} + 2k\pi \end{aligned}

Example 3:

\begin{aligned} \tan (x) = \sqrt {3} \end{aligned}

Solution : We know that the $\tan {\frac{\pi}{3}} = \sqrt{3}$ and therefore

\begin{aligned} \tan (x) = \tan ( \frac{\pi}{3} )\mathop \to \limits^{\color{red}{formula3}} &x = \frac{\pi}{3} + k\pi \end{aligned}

Example 4:

\begin{aligned} 2\sin(2x) - 1 = 0 \end{aligned}

Solution:

Step 1: To solve for $x$ , you must first isolate the sine term.

\begin{aligned} 2\sin(2x) - 1 &= 0 \\ 2\sin(2x) &= 1 \\ \sin(2x) &= \frac{1}{2} \end{aligned}

Step 2: We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and therefore

\begin{aligned} \sin (2x) = \sin ( \frac{\pi}{6} )\mathop \to \limits^{\color{red}{formula1}} &2x = \frac{\pi}{6} + k\pi \hspace{0.5cm} \text{or} \hspace{0.5cm} \to x = \frac{\pi}{12} + k\pi \hspace{0.5cm} \text{or} \\ &2x = \frac {5\pi}{6} + 2k\pi \hspace{1.7cm} x = \frac{5\pi}{12} + k\pi \end{aligned}

Example 5:

3 (\tan(x))^2 = 1

Solution:

Step 1: To solve for $x$ , firstly, you must isolate the tangent term.

\begin{aligned} 3 (\tan(x))^2 &= 1 \\ (\tan(x))^2 &= \frac{1}{3} \\ \tan (x) &= \color{red}{\pm} \sqrt {\frac{1}{3}} \\ \tan (x) &= \pm \frac{1}{\sqrt{3}} \frac{\sqrt{3}}{\sqrt{3}} \\ \tan (x) &= \pm \frac{\sqrt{3}}{3} \end{aligned}

Step 2: We know that $\tan (\frac{\pi}{6}) = \frac{\sqrt{3}}{3}$ and $\tan (-\frac{\pi}{6}) = - \frac{\sqrt{3}}{3}$ , therefore

\begin{aligned} \tan (x) = \tan (\frac{\pi }{6}) &\Rightarrow x = \frac{\pi }{6} + k\pi \ \text{and} \\ \tan (x) = \tan (-\frac{\pi }{6}) &\Rightarrow x = -\frac{\pi }{6} + k\pi \end{aligned}

Example 6:

2 \cos(2x - \frac{\pi}{3}) = 1

Solution

Step 1: To solve for $x$ , firstly, you must isolate the cosine term.

\begin{aligned} 2\cos(2x - \frac{\pi}{3}) &= 1 \\ \cos (2x - \frac{\pi}{3}) &= \frac{1}{2} \end{aligned}

Step 2: We know that $\cos (\frac{\pi}{3}) = \frac{1}{2}$ , therefore

\begin{aligned} \cos (2x - \frac{\pi }{3}) = \cos (\frac{\pi }{3})\mathop \to \limits^{\color{red}{formula2}} &2x - \frac{\pi }{3} + 2k\pi \ \text{or} \to 2x = \frac{\pi}{6} + 2k\pi \ \text{or} \to x = \frac{\pi}{12}+k\pi \ \text{or} \\ &2x - \frac{\pi}{3} = -\frac{\pi}{3} + 2k\pi \hspace{1cm} 2x = 0 + 2k\pi \hspace{1cm} x = k\pi \end{aligned}

Algebra

It is a mathematical fact that fifty percent of all doctors graduate in the bottom half of their class.

Author Unknown

There are things which seem incredible to most men who have not studied Mathematics.

Archimedes of Syracus