Solve sin(x)² = 1/4
Trigonometric equation, worked out line by line the way a teacher would write it.
Answer
| General solution | x = π/6 + 2π·n; x = 5π/6 + 2π·n; x = 7π/6 + 2π·n; x = 11π/6 + 2π·n (n any integer) |
Step-by-step solution
16 steps-
1 Given\sin^{2}{\left(x \right)} = \frac{1}{4}
Solve for x.
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2 Substitute t = sin(x)
Treat sin(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} - \frac{1}{4} = 0 -
3 Multiply both sides by 4 to clear the fractions
4 is the least common multiple of the denominators. Multiplying every term on both sides by it cancels each fraction, so the rest of the work uses whole numbers. The equation stays balanced because both sides were multiplied by the same number.
4 t^{2} - 1 = 0 -
4 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 = 4, b = 0 and c = -1, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 4,\quad b = 0,\quad c = -1 -
5 There is no t term, so isolate t²t^{2} = \frac{1}{4}
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6 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 \frac{1}{2} -
7 Solutionst = \frac{1}{2}\quad \text{or} \quad t = - \frac{1}{2}
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8 Back-substitute: sin(x) = 1/2\sin{\left(x \right)} = \frac{1}{2}
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9 Reference angle: the acute angle α with sin 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 = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.5235987902 -
10 Sine is positive in quadrants I and II: u = α or u = π − 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}{6}\quad \text{or} \quad x = \frac{5 \pi}{6} -
11 Back-substitute: sin(x) = -1/2\sin{\left(x \right)} = - \frac{1}{2}
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12 Reference angle: the acute angle α with sin α = 1/2\alpha = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.5235987902
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13 Sine is negative in quadrants III and IV: u = π + α or u = 2π − αx = \frac{7 \pi}{6}\quad \text{or} \quad x = \frac{11 \pi}{6}
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14 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}{6} + 2 \pi n\quad \text{or} \quad x = \frac{5 \pi}{6} + 2 \pi n\quad \text{or} \quad x = \frac{7 \pi}{6} + 2 \pi n\quad \text{or} \quad x = \frac{11 \pi}{6} + 2 \pi n,\quad n \in \mathbb{Z} -
15 The solutions in one turn, 0 ≤ x < 2πx = \frac{\pi}{6},\; \frac{5 \pi}{6},\; \frac{7 \pi}{6},\; \frac{11 \pi}{6}
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16 Check\begin{aligned}x = \frac{\pi}{6}:\quad \frac{1}{4} = \frac{1}{4}\quad\checkmark\\ x = \frac{5 \pi}{6}:\quad \frac{1}{4} = \frac{1}{4}\quad\checkmark\\ x = \frac{7 \pi}{6}:\quad \frac{1}{4} = \frac{1}{4}\quad\checkmark\\ x = \frac{11 \pi}{6}:\quad \frac{1}{4} = \frac{1}{4}\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| pi/6 | 1/4 | 1/4 | ✓ |
| 5pi/6 | 1/4 | 1/4 | ✓ |
| 7pi/6 | 1/4 | 1/4 | ✓ |
| 11pi/6 | 1/4 | 1/4 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.