Solve tan(x)² = 3
Trigonometric equation, worked out line by line the way a teacher would write it.
Answer
| General solution | x = π/3 + π·n; x = 2π/3 + π·n (n any integer) |
Step-by-step solution
15 steps-
1 Given\tan^{2}{\left(x \right)} = 3
Solve for x.
-
2 Substitute t = tan(x)
Treat tan(x) as a single unknown t: the equation becomes an ordinary polynomial in t. Solve for t, then find the angles that have each value.
t^{2} - 3 = 0 -
3 This is a quadratic in standard form at² + bt + c = 0
Every quadratic equation can be arranged as at² + bt + c = 0. Reading off a = 1, b = 0 and c = -3, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 0,\quad c = -3 -
4 There is no t term, so isolate t²t^{2} = 3
-
5 Take the square root of both sides — remember both signs
A number and its negative give the same result when raised to an even power: 3² = 9 and (−3)² = 9. So if something squared equals k, that something is √k or −√k. Forgetting the minus sign loses a solution.
t = \pm \sqrt{3} -
6 Solutionst = \sqrt{3}\quad \text{or} \quad t = - \sqrt{3}
-
7 Back-substitute: tan(x) = √3\tan{\left(x \right)} = \sqrt{3}
-
8 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 -
9 Tangent is positive in quadrant I 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{\pi}{3} -
10 Back-substitute: tan(x) = -√3\tan{\left(x \right)} = - \sqrt{3}
-
11 Reference angle: the acute angle α with tan α = √3\alpha = \arctan\left(\sqrt{3}\right) = \frac{\pi}{3} \approx 1.04719758
-
12 Tangent is negative in quadrant II and repeats every half turnx = \frac{2 \pi}{3}
-
13 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{\pi}{3} + \pi n\quad \text{or} \quad x = \frac{2 \pi}{3} + \pi n,\quad n \in \mathbb{Z} -
14 The solutions in one turn, 0 ≤ x < 2πx = \frac{\pi}{3},\; \frac{2 \pi}{3},\; \frac{4 \pi}{3},\; \frac{5 \pi}{3}
-
15 Check\begin{aligned}x = \frac{\pi}{3}:\quad 3 = 3\quad\checkmark\\ x = \frac{2 \pi}{3}:\quad 3 = 3\quad\checkmark\\ x = \frac{4 \pi}{3}:\quad 3 = 3\quad\checkmark\\ x = \frac{5 \pi}{3}:\quad 3 = 3\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| pi/3 | 3 | 3 | ✓ |
| 2pi/3 | 3 | 3 | ✓ |
| 4pi/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.