Solve 2cos(x) − 1 = 0
Trigonometric equation, worked out line by line the way a teacher would write it.
Answer
| General solution | x = π/3 + 2π·n; x = 5π/3 + 2π·n (n any integer) |
Step-by-step solution
7 steps-
1 Given2 \cos{\left(x \right)} - 1 = 0
Solve for x.
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2 Isolate cos(x)\cos{\left(x \right)} = \frac{1}{2}
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3 Reference angle: the acute angle α with cos α = 1/2
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 = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \approx 1.04719758 -
4 Cosine is positive in quadrants I and IV: u = α or u = 2π − α
In one full turn, sine and cosine reach each value between −1 and 1 twice. The signs follow the quadrants: everything is positive in quadrant I, only sine in II, only tangent in III and only cosine in IV.
x = \frac{\pi}{3}\quad \text{or} \quad x = \frac{5 \pi}{3} -
5 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} + 2 \pi n\quad \text{or} \quad x = \frac{5 \pi}{3} + 2 \pi n,\quad n \in \mathbb{Z} -
6 The solutions in one turn, 0 ≤ x < 2πx = \frac{\pi}{3},\; \frac{5 \pi}{3}
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7 Check\begin{aligned}x = \frac{\pi}{3}:\quad 0 = 0\quad\checkmark\\ x = \frac{5 \pi}{3}:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| pi/3 | 0 | 0 | ✓ |
| 5pi/3 | 0 | 0 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.