Solve tan(x) = −√(3)
Trigonometric equation, worked out line by line the way a teacher would write it.
Answer
| General solution | x = 2π/3 + π·n (n any integer) |
Step-by-step solution
6 steps-
1 Given\tan{\left(x \right)} = - \sqrt{3}
Solve for x.
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2 Reference angle: the acute angle α with tan α = √3
The reference angle α is the acute angle (between 0 and 90°) whose sine, cosine or tangent has this value, ignoring the sign. The other angles with the same value are built from it in the next step.
\alpha = \arctan\left(\sqrt{3}\right) = \frac{\pi}{3} \approx 1.04719758 -
3 Tangent is negative in quadrant II and repeats every half turn
Tangent repeats every half turn (180° or π), so one angle per half turn is enough. It is positive in quadrants I and III and negative in quadrants II and IV.
x = \frac{2 \pi}{3} -
4 General solution (n is any integer)
Sine and cosine repeat every full turn (360° or 2π) and tangent every half turn, so adding any whole number n of periods gives another solution. n can be any integer, positive, negative or 0.
x = \frac{2 \pi}{3} + \pi n,\quad n \in \mathbb{Z} -
5 The solutions in one turn, 0 ≤ x < 2πx = \frac{2 \pi}{3},\; \frac{5 \pi}{3}
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6 Check\begin{aligned}x = \frac{2 \pi}{3}:\quad - \sqrt{3} = - \sqrt{3}\quad\checkmark\\ x = \frac{5 \pi}{3}:\quad - \sqrt{3} = - \sqrt{3}\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| 2pi/3 | −√3 | −√3 | ✓ |
| 5pi/3 | −√3 | −√3 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.