Trig Equations worksheet 5.1 featuring ten trigonometric equations to solve within the interval 0 ≤ x < 2π.
Trig Equations worksheet 5.1 with ten problems to solve for 0 ≤ x < 2π, including equations involving sin, cos, tan, cot, csc, and sec functions.
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Step-by-step solution for: Trig Equations worksheet 5.1 Name: Solve for 0≤x
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Show Answer Key & Explanations
Step-by-step solution for: Trig Equations worksheet 5.1 Name: Solve for 0≤x
Let’s solve each trig equation step by step for $0 \le x < 2\pi$. We’ll go one at a time.
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1. $\sin x = \frac{1}{2}$
We know sine is positive in Quadrants I and II.
Reference angle: $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
So solutions:
- QI: $x = \frac{\pi}{6}$
- QII: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
✔ Final Answer for #1: $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}}$
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2. $\tan x = 1$
Tangent is positive in QI and QIII.
Reference angle: $\tan^{-1}(1) = \frac{\pi}{4}$
Solutions:
- QI: $x = \frac{\pi}{4}$
- QIII: $x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$
✔ Final Answer for #2: $\boxed{\frac{\pi}{4}, \frac{5\pi}{4}}$
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3. $2\sin x + \sqrt{3} = 0$
Subtract $\sqrt{3}$:
$2\sin x = -\sqrt{3}$
Divide by 2:
$\sin x = -\frac{\sqrt{3}}{2}$
Sine is negative in QIII and QIV.
Reference angle: $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$
Solutions:
- QIII: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
- QIV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
✔ Final Answer for #3: $\boxed{\frac{4\pi}{3}, \frac{5\pi}{3}}$
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4. $2\cos x + \sqrt{3} = 0$
Subtract $\sqrt{3}$:
$2\cos x = -\sqrt{3}$
Divide by 2:
$\cos x = -\frac{\sqrt{3}}{2}$
Cosine is negative in QII and QIII.
Reference angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
Solutions:
- QII: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
- QIII: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$
✔ Final Answer for #4: $\boxed{\frac{5\pi}{6}, \frac{7\pi}{6}}$
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5. $2\cos x - \sqrt{3} = 0$
Add $\sqrt{3}$:
$2\cos x = \sqrt{3}$
Divide by 2:
$\cos x = \frac{\sqrt{3}}{2}$
Cosine is positive in QI and QIV.
Reference angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
Solutions:
- QI: $x = \frac{\pi}{6}$
- QIV: $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$
✔ Final Answer for #5: $\boxed{\frac{\pi}{6}, \frac{11\pi}{6}}$
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6. $\tan x + 1 = 0$
Subtract 1:
$\tan x = -1$
Tangent is negative in QII and QIV.
Reference angle: $\tan^{-1}(1) = \frac{\pi}{4}$
Solutions:
- QII: $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
- QIV: $x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$
✔ Final Answer for #6: $\boxed{\frac{3\pi}{4}, \frac{7\pi}{4}}$
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7. $\sqrt{3}\cot x + 1 = 0$
Subtract 1:
$\sqrt{3}\cot x = -1$
Divide by $\sqrt{3}$:
$\cot x = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$ (rationalized)
Cotangent is reciprocal of tangent, so same sign behavior.
$\cot x = -\frac{\sqrt{3}}{3}$ → $\tan x = -\sqrt{3}$
Tangent is negative in QII and QIV.
Reference angle: $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$
Solutions:
- QII: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
- QIV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
✔ Final Answer for #7: $\boxed{\frac{2\pi}{3}, \frac{5\pi}{3}}$
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8. $\csc^2 x = 1$
Take square root:
$\csc x = \pm 1$
But $\csc x = \frac{1}{\sin x}$, so:
If $\csc x = 1$, then $\sin x = 1$ → $x = \frac{\pi}{2}$
If $\csc x = -1$, then $\sin x = -1$ → $x = \frac{3\pi}{2}$
Check domain: both are within $[0, 2\pi)$
✔ Final Answer for #8: $\boxed{\frac{\pi}{2}, \frac{3\pi}{2}}$
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9. $\sec^2 x - 1 = 0$
Add 1:
$\sec^2 x = 1$
Take square root:
$\sec x = \pm 1$
$\sec x = \frac{1}{\cos x}$, so:
If $\sec x = 1$ → $\cos x = 1$ → $x = 0$
If $\sec x = -1$ → $\cos x = -1$ → $x = \pi$
Both in range.
✔ Final Answer for #9: $\boxed{0, \pi}$
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10. $4\sec^2 x = 8$
Divide by 4:
$\sec^2 x = 2$
Take square root:
$\sec x = \pm \sqrt{2}$
So $\cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$
Now find all angles where cosine is $\frac{\sqrt{2}}{2}$ or $-\frac{\sqrt{2}}{2}$
Reference angle: $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$
Cases:
- $\cos x = \frac{\sqrt{2}}{2}$ → QI and QIV: $x = \frac{\pi}{4}, \frac{7\pi}{4}$
- $\cos x = -\frac{\sqrt{2}}{2}$ → QII and QIII: $x = \frac{3\pi}{4}, \frac{5\pi}{4}$
All four are valid.
✔ Final Answer for #10: $\boxed{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}}$
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Final Answer:
1. $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}}$
2. $\boxed{\frac{\pi}{4}, \frac{5\pi}{4}}$
3. $\boxed{\frac{4\pi}{3}, \frac{5\pi}{3}}$
4. $\boxed{\frac{5\pi}{6}, \frac{7\pi}{6}}$
5. $\boxed{\frac{\pi}{6}, \frac{11\pi}{6}}$
6. $\boxed{\frac{3\pi}{4}, \frac{7\pi}{4}}$
7. $\boxed{\frac{2\pi}{3}, \frac{5\pi}{3}}$
8. $\boxed{\frac{\pi}{2}, \frac{3\pi}{2}}$
9. $\boxed{0, \pi}$
10. $\boxed{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}}$
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1. $\sin x = \frac{1}{2}$
We know sine is positive in Quadrants I and II.
Reference angle: $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
So solutions:
- QI: $x = \frac{\pi}{6}$
- QII: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
✔ Final Answer for #1: $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}}$
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2. $\tan x = 1$
Tangent is positive in QI and QIII.
Reference angle: $\tan^{-1}(1) = \frac{\pi}{4}$
Solutions:
- QI: $x = \frac{\pi}{4}$
- QIII: $x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$
✔ Final Answer for #2: $\boxed{\frac{\pi}{4}, \frac{5\pi}{4}}$
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3. $2\sin x + \sqrt{3} = 0$
Subtract $\sqrt{3}$:
$2\sin x = -\sqrt{3}$
Divide by 2:
$\sin x = -\frac{\sqrt{3}}{2}$
Sine is negative in QIII and QIV.
Reference angle: $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$
Solutions:
- QIII: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
- QIV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
✔ Final Answer for #3: $\boxed{\frac{4\pi}{3}, \frac{5\pi}{3}}$
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4. $2\cos x + \sqrt{3} = 0$
Subtract $\sqrt{3}$:
$2\cos x = -\sqrt{3}$
Divide by 2:
$\cos x = -\frac{\sqrt{3}}{2}$
Cosine is negative in QII and QIII.
Reference angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
Solutions:
- QII: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
- QIII: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$
✔ Final Answer for #4: $\boxed{\frac{5\pi}{6}, \frac{7\pi}{6}}$
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5. $2\cos x - \sqrt{3} = 0$
Add $\sqrt{3}$:
$2\cos x = \sqrt{3}$
Divide by 2:
$\cos x = \frac{\sqrt{3}}{2}$
Cosine is positive in QI and QIV.
Reference angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
Solutions:
- QI: $x = \frac{\pi}{6}$
- QIV: $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$
✔ Final Answer for #5: $\boxed{\frac{\pi}{6}, \frac{11\pi}{6}}$
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6. $\tan x + 1 = 0$
Subtract 1:
$\tan x = -1$
Tangent is negative in QII and QIV.
Reference angle: $\tan^{-1}(1) = \frac{\pi}{4}$
Solutions:
- QII: $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
- QIV: $x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$
✔ Final Answer for #6: $\boxed{\frac{3\pi}{4}, \frac{7\pi}{4}}$
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7. $\sqrt{3}\cot x + 1 = 0$
Subtract 1:
$\sqrt{3}\cot x = -1$
Divide by $\sqrt{3}$:
$\cot x = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$ (rationalized)
Cotangent is reciprocal of tangent, so same sign behavior.
$\cot x = -\frac{\sqrt{3}}{3}$ → $\tan x = -\sqrt{3}$
Tangent is negative in QII and QIV.
Reference angle: $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$
Solutions:
- QII: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
- QIV: $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
✔ Final Answer for #7: $\boxed{\frac{2\pi}{3}, \frac{5\pi}{3}}$
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8. $\csc^2 x = 1$
Take square root:
$\csc x = \pm 1$
But $\csc x = \frac{1}{\sin x}$, so:
If $\csc x = 1$, then $\sin x = 1$ → $x = \frac{\pi}{2}$
If $\csc x = -1$, then $\sin x = -1$ → $x = \frac{3\pi}{2}$
Check domain: both are within $[0, 2\pi)$
✔ Final Answer for #8: $\boxed{\frac{\pi}{2}, \frac{3\pi}{2}}$
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9. $\sec^2 x - 1 = 0$
Add 1:
$\sec^2 x = 1$
Take square root:
$\sec x = \pm 1$
$\sec x = \frac{1}{\cos x}$, so:
If $\sec x = 1$ → $\cos x = 1$ → $x = 0$
If $\sec x = -1$ → $\cos x = -1$ → $x = \pi$
Both in range.
✔ Final Answer for #9: $\boxed{0, \pi}$
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10. $4\sec^2 x = 8$
Divide by 4:
$\sec^2 x = 2$
Take square root:
$\sec x = \pm \sqrt{2}$
So $\cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$
Now find all angles where cosine is $\frac{\sqrt{2}}{2}$ or $-\frac{\sqrt{2}}{2}$
Reference angle: $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$
Cases:
- $\cos x = \frac{\sqrt{2}}{2}$ → QI and QIV: $x = \frac{\pi}{4}, \frac{7\pi}{4}$
- $\cos x = -\frac{\sqrt{2}}{2}$ → QII and QIII: $x = \frac{3\pi}{4}, \frac{5\pi}{4}$
All four are valid.
✔ Final Answer for #10: $\boxed{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}}$
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Final Answer:
1. $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}}$
2. $\boxed{\frac{\pi}{4}, \frac{5\pi}{4}}$
3. $\boxed{\frac{4\pi}{3}, \frac{5\pi}{3}}$
4. $\boxed{\frac{5\pi}{6}, \frac{7\pi}{6}}$
5. $\boxed{\frac{\pi}{6}, \frac{11\pi}{6}}$
6. $\boxed{\frac{3\pi}{4}, \frac{7\pi}{4}}$
7. $\boxed{\frac{2\pi}{3}, \frac{5\pi}{3}}$
8. $\boxed{\frac{\pi}{2}, \frac{3\pi}{2}}$
9. $\boxed{0, \pi}$
10. $\boxed{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}}$
Parent Tip: Review the logic above to help your child master the concept of solving trig equations worksheet.