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Trigonometry: (lesson 2 of 3)

Trigonometric Equations

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}

Basic trigonometric equations:

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}