Solve cos(x) = −√(3)/2 in degrees
Trigonometric equation, worked out line by line the way a teacher would write it.
Answer
| General solution | x = 150° + 360°·n; x = 210° + 360°·n (n any integer) |
Step-by-step solution
6 steps-
1 Given\cos{\left(x \right)} = - \frac{\sqrt{3}}{2}
Solve for x. Angles are in degrees.
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2 Reference angle: the acute angle α with cos α = √3/2
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{\sqrt{3}}{2}\right) = 30^{\circ} -
3 Cosine is negative in quadrants II and III: u = 180° − α or u = 180° + 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 = 150^{\circ}\quad \text{or} \quad x = 210^{\circ} -
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 = 150^{\circ} + 360^{\circ} n\quad \text{or} \quad x = 210^{\circ} + 360^{\circ} n,\quad n \in \mathbb{Z} -
5 The solutions in one turn, 0 ≤ x < 360°x = 150^{\circ},\; 210^{\circ}
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6 Check\begin{aligned}x = 150:\quad - \frac{\sqrt{3}}{2} = - \frac{\sqrt{3}}{2}\quad\checkmark\\ x = 210:\quad - \frac{\sqrt{3}}{2} = - \frac{\sqrt{3}}{2}\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| 150 | −√3/2 | −√3/2 | ✓ |
| 210 | −√3/2 | −√3/2 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.