CLEP College Mathematics Worksheets: FREE & Printable - Effortless - Free Printable
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Step-by-step solution for: CLEP College Mathematics Worksheets: FREE & Printable - Effortless
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Show Answer Key & Explanations
Step-by-step solution for: CLEP College Mathematics Worksheets: FREE & Printable - Effortless
Let’s solve each equation one by one for $0^\circ \le x \le 360^\circ$, rounding answers to the nearest tenth.
---
1) sin x = 1
→ The sine function equals 1 at 90° only in this range.
✔ Answer: 90.0°
---
2) tan x = 1
→ Tangent is positive in Q1 and Q3.
→ Reference angle: arctan(1) = 45°
→ So, x = 45° and 180° + 45° = 225°
✔ Answers: 45.0°, 225.0°
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3) cos x = 1
→ Cosine equals 1 at 0° and also at 360° (since it's included).
✔ Answers: 0.0°, 360.0°
---
4) sin x = 0.5
→ Sine is positive in Q1 and Q2.
→ Reference angle: arcsin(0.5) = 30°
→ So, x = 30° and 180° - 30° = 150°
✔ Answers: 30.0°, 150.0°
---
5) cos x = 0.6
→ Cosine is positive in Q1 and Q4.
→ Reference angle: arccos(0.6) ≈ 53.13° → round to 53.1°
→ So, x = 53.1° and 360° - 53.1° = 306.9°
✔ Answers: 53.1°, 306.9°
---
6) tan x = 0.2
→ Tangent is positive in Q1 and Q3.
→ Reference angle: arctan(0.2) ≈ 11.31° → round to 11.3°
→ So, x = 11.3° and 180° + 11.3° = 191.3°
✔ Answers: 11.3°, 191.3°
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7) 7 cos x = 3
→ Divide both sides: cos x = 3/7 ≈ 0.4286
→ arccos(0.4286) ≈ 64.62° → round to 64.6°
→ Cosine positive in Q1 and Q4 → x = 64.6° and 360° - 64.6° = 295.4°
✔ Answers: 64.6°, 295.4°
---
8) 2 tan x = ¼
→ tan x = (1/4)/2 = 1/8 = 0.125
→ arctan(0.125) ≈ 7.125° → round to 7.1°
→ Tangent positive in Q1 and Q3 → x = 7.1° and 180° + 7.1° = 187.1°
✔ Answers: 7.1°, 187.1°
---
9) 8 + 3 sin x = 10
→ Subtract 8: 3 sin x = 2 → sin x = 2/3 ≈ 0.6667
→ arcsin(0.6667) ≈ 41.81° → round to 41.8°
→ Sine positive in Q1 and Q2 → x = 41.8° and 180° - 41.8° = 138.2°
✔ Answers: 41.8°, 138.2°
---
10) tan x = -1
→ Tangent negative in Q2 and Q4.
→ Reference angle: arctan(1) = 45°
→ So, x = 180° - 45° = 135° and 360° - 45° = 315°
✔ Answers: 135.0°, 315.0°
---
11) sin x = -0.9
→ Sine negative in Q3 and Q4.
→ Reference angle: arcsin(0.9) ≈ 64.16° → round to 64.2°
→ So, x = 180° + 64.2° = 244.2° and 360° - 64.2° = 295.8°
✔ Answers: 244.2°, 295.8°
---
12) tan x = -0.1
→ Tangent negative in Q2 and Q4.
→ Reference angle: arctan(0.1) ≈ 5.71° → round to 5.7°
→ So, x = 180° - 5.7° = 174.3° and 360° - 5.7° = 354.3°
✔ Answers: 174.3°, 354.3°
---
13) 7 + 8 sin x = 4
→ Subtract 7: 8 sin x = -3 → sin x = -3/8 = -0.375
→ Reference angle: arcsin(0.375) ≈ 22.02° → round to 22.0°
→ Sine negative in Q3 and Q4 → x = 180° + 22.0° = 202.0° and 360° - 22.0° = 338.0°
✔ Answers: 202.0°, 338.0°
---
14) sin 2x = 0.2886
→ Let θ = 2x → sin θ = 0.2886 → θ ≈ arcsin(0.2886) ≈ 16.77° → round to 16.8°
→ But since 0° ≤ x ≤ 360°, then 0° ≤ 2x ≤ 720° → so we need all solutions for θ in [0°, 720°]
Solutions for θ:
- Q1: 16.8°
- Q2: 180° - 16.8° = 163.2°
- Add 360°: 16.8° + 360° = 376.8°, 163.2° + 360° = 523.2°
Now divide by 2 to get x:
- x = 16.8° / 2 = 8.4°
- x = 163.2° / 2 = 81.6°
- x = 376.8° / 2 = 188.4°
- x = 523.2° / 2 = 261.6°
✔ Answers: 8.4°, 81.6°, 188.4°, 261.6°
---
15) cos 3x = -0.3321
→ Let θ = 3x → cos θ = -0.3321 → reference angle: arccos(0.3321) ≈ 70.6° → so angles where cosine is negative: Q2 and Q3
θ = 180° - 70.6° = 109.4°
θ = 180° + 70.6° = 250.6°
But 0° ≤ x ≤ 360° → 0° ≤ 3x ≤ 1080° → so find all θ in [0°, 1080°] that satisfy cos θ = -0.3321
Add 360° repeatedly:
First cycle:
- 109.4°, 250.6°
Second cycle (+360):
- 109.4° + 360° = 469.4°
- 250.6° + 360° = 610.6°
Third cycle (+720):
- 109.4° + 720° = 829.4°
- 250.6° + 720° = 970.6°
Now divide each by 3 to get x:
- 109.4° / 3 ≈ 36.5°
- 250.6° / 3 ≈ 83.5°
- 469.4° / 3 ≈ 156.5°
- 610.6° / 3 ≈ 203.5°
- 829.4° / 3 ≈ 276.5°
- 970.6° / 3 ≈ 323.5°
✔ Answers: 36.5°, 83.5°, 156.5°, 203.5°, 276.5°, 323.5°
---
16) sin(x + 20°) = 0.6551
→ Let θ = x + 20° → sin θ = 0.6551 → θ ≈ arcsin(0.6551) ≈ 40.9° (Q1), and 180° - 40.9° = 139.1° (Q2)
So:
- x + 20° = 40.9° → x = 20.9°
- x + 20° = 139.1° → x = 119.1°
Also, since sine repeats every 360°, add 360° to θ:
- θ = 40.9° + 360° = 400.9° → x = 380.9° → too big (>360°)
- θ = 139.1° + 360° = 499.1° → x = 479.1° → too big
So only two solutions.
✔ Answers: 20.9°, 119.1°
---
17) tan(x - 15°) = -0.9128
→ Let θ = x - 15° → tan θ = -0.9128 → reference angle: arctan(0.9128) ≈ 42.4°
Tangent negative in Q2 and Q4:
θ = 180° - 42.4° = 137.6°
θ = 360° - 42.4° = 317.6°
Also, tangent has period 180°, so add 180° to each:
θ = 137.6° + 180° = 317.6° → already have
θ = 317.6° + 180° = 497.6° → check if within range?
Since 0° ≤ x ≤ 360° → θ = x - 15° → -15° ≤ θ ≤ 345°
Wait — let’s be careful.
Actually, θ can go from -15° to 345°, but tangent is periodic with 180°, so we should consider all θ such that tan θ = -0.9128 and θ ∈ [-15°, 345°]
We have:
θ₁ = 137.6° → x = 137.6° + 15° = 152.6°
θ₂ = 317.6° → x = 317.6° + 15° = 332.6°
What about subtracting 180°? θ = 137.6° - 180° = -42.4° → which is ≥ -15°? No, -42.4 < -15 → invalid.
θ = 317.6° - 180° = 137.6° → already have.
So only two solutions.
✔ Answers: 152.6°, 332.6°
---
18) cos(2x + 33°) = 0.306
→ Let θ = 2x + 33° → cos θ = 0.306 → θ ≈ arccos(0.306) ≈ 72.2° (Q1), and 360° - 72.2° = 287.8° (Q4)
Now, 0° ≤ x ≤ 360° → 33° ≤ θ ≤ 2*360° + 33° = 753°
So find all θ in [33°, 753°] where cos θ = 0.306
Start with base solutions:
θ₁ = 72.2°
θ₂ = 287.8°
Add 360°:
θ₃ = 72.2° + 360° = 432.2°
θ₄ = 287.8° + 360° = 647.8°
Add another 360°? 72.2° + 720° = 792.2° > 753° → stop.
So four values: 72.2°, 287.8°, 432.2°, 647.8°
Now solve for x:
θ = 2x + 33° → 2x = θ - 33° → x = (θ - 33°)/2
- x₁ = (72.2 - 33)/2 = 39.2/2 = 19.6°
- x₂ = (287.8 - 33)/2 = 254.8/2 = 127.4°
- x₃ = (432.2 - 33)/2 = 399.2/2 = 199.6°
- x₄ = (647.8 - 33)/2 = 614.8/2 = 307.4°
✔ Answers: 19.6°, 127.4°, 199.6°, 307.4°
---
19) tan²x = ¼
→ Take square root: tan x = ±½
Case 1: tan x = 0.5 → as in #6: x ≈ 26.6° and 206.6°? Wait no — earlier for 0.2 it was 11.3°, now 0.5:
arctan(0.5) ≈ 26.565° → round to 26.6°
So solutions: 26.6°, 180° + 26.6° = 206.6°
Case 2: tan x = -0.5 → reference angle 26.6°, negative in Q2 and Q4:
x = 180° - 26.6° = 153.4°
x = 360° - 26.6° = 333.4°
✔ Answers: 26.6°, 153.4°, 206.6°, 333.4°
---
20) cos²x = ½
→ cos x = ±√(½) = ±√2/2 ≈ ±0.7071
Case 1: cos x = √2/2 → x = 45°, 315°
Case 2: cos x = -√2/2 → x = 135°, 225°
All exact values.
✔ Answers: 45.0°, 135.0°, 225.0°, 315.0°
---
Final Answer:
1) 90.0°
2) 45.0°, 225.0°
3) 0.0°, 360.0°
4) 30.0°, 150.0°
5) 53.1°, 306.9°
6) 11.3°, 191.3°
7) 64.6°, 295.4°
8) 7.1°, 187.1°
9) 41.8°, 138.2°
10) 135.0°, 315.0°
11) 244.2°, 295.8°
12) 174.3°, 354.3°
13) 202.0°, 338.0°
14) 8.4°, 81.6°, 188.4°, 261.6°
15) 36.5°, 83.5°, 156.5°, 203.5°, 276.5°, 323.5°
16) 20.9°, 119.1°
17) 152.6°, 332.6°
18) 19.6°, 127.4°, 199.6°, 307.4°
19) 26.6°, 153.4°, 206.6°, 333.4°
20) 45.0°, 135.0°, 225.0°, 315.0°
---
1) sin x = 1
→ The sine function equals 1 at 90° only in this range.
✔ Answer: 90.0°
---
2) tan x = 1
→ Tangent is positive in Q1 and Q3.
→ Reference angle: arctan(1) = 45°
→ So, x = 45° and 180° + 45° = 225°
✔ Answers: 45.0°, 225.0°
---
3) cos x = 1
→ Cosine equals 1 at 0° and also at 360° (since it's included).
✔ Answers: 0.0°, 360.0°
---
4) sin x = 0.5
→ Sine is positive in Q1 and Q2.
→ Reference angle: arcsin(0.5) = 30°
→ So, x = 30° and 180° - 30° = 150°
✔ Answers: 30.0°, 150.0°
---
5) cos x = 0.6
→ Cosine is positive in Q1 and Q4.
→ Reference angle: arccos(0.6) ≈ 53.13° → round to 53.1°
→ So, x = 53.1° and 360° - 53.1° = 306.9°
✔ Answers: 53.1°, 306.9°
---
6) tan x = 0.2
→ Tangent is positive in Q1 and Q3.
→ Reference angle: arctan(0.2) ≈ 11.31° → round to 11.3°
→ So, x = 11.3° and 180° + 11.3° = 191.3°
✔ Answers: 11.3°, 191.3°
---
7) 7 cos x = 3
→ Divide both sides: cos x = 3/7 ≈ 0.4286
→ arccos(0.4286) ≈ 64.62° → round to 64.6°
→ Cosine positive in Q1 and Q4 → x = 64.6° and 360° - 64.6° = 295.4°
✔ Answers: 64.6°, 295.4°
---
8) 2 tan x = ¼
→ tan x = (1/4)/2 = 1/8 = 0.125
→ arctan(0.125) ≈ 7.125° → round to 7.1°
→ Tangent positive in Q1 and Q3 → x = 7.1° and 180° + 7.1° = 187.1°
✔ Answers: 7.1°, 187.1°
---
9) 8 + 3 sin x = 10
→ Subtract 8: 3 sin x = 2 → sin x = 2/3 ≈ 0.6667
→ arcsin(0.6667) ≈ 41.81° → round to 41.8°
→ Sine positive in Q1 and Q2 → x = 41.8° and 180° - 41.8° = 138.2°
✔ Answers: 41.8°, 138.2°
---
10) tan x = -1
→ Tangent negative in Q2 and Q4.
→ Reference angle: arctan(1) = 45°
→ So, x = 180° - 45° = 135° and 360° - 45° = 315°
✔ Answers: 135.0°, 315.0°
---
11) sin x = -0.9
→ Sine negative in Q3 and Q4.
→ Reference angle: arcsin(0.9) ≈ 64.16° → round to 64.2°
→ So, x = 180° + 64.2° = 244.2° and 360° - 64.2° = 295.8°
✔ Answers: 244.2°, 295.8°
---
12) tan x = -0.1
→ Tangent negative in Q2 and Q4.
→ Reference angle: arctan(0.1) ≈ 5.71° → round to 5.7°
→ So, x = 180° - 5.7° = 174.3° and 360° - 5.7° = 354.3°
✔ Answers: 174.3°, 354.3°
---
13) 7 + 8 sin x = 4
→ Subtract 7: 8 sin x = -3 → sin x = -3/8 = -0.375
→ Reference angle: arcsin(0.375) ≈ 22.02° → round to 22.0°
→ Sine negative in Q3 and Q4 → x = 180° + 22.0° = 202.0° and 360° - 22.0° = 338.0°
✔ Answers: 202.0°, 338.0°
---
14) sin 2x = 0.2886
→ Let θ = 2x → sin θ = 0.2886 → θ ≈ arcsin(0.2886) ≈ 16.77° → round to 16.8°
→ But since 0° ≤ x ≤ 360°, then 0° ≤ 2x ≤ 720° → so we need all solutions for θ in [0°, 720°]
Solutions for θ:
- Q1: 16.8°
- Q2: 180° - 16.8° = 163.2°
- Add 360°: 16.8° + 360° = 376.8°, 163.2° + 360° = 523.2°
Now divide by 2 to get x:
- x = 16.8° / 2 = 8.4°
- x = 163.2° / 2 = 81.6°
- x = 376.8° / 2 = 188.4°
- x = 523.2° / 2 = 261.6°
✔ Answers: 8.4°, 81.6°, 188.4°, 261.6°
---
15) cos 3x = -0.3321
→ Let θ = 3x → cos θ = -0.3321 → reference angle: arccos(0.3321) ≈ 70.6° → so angles where cosine is negative: Q2 and Q3
θ = 180° - 70.6° = 109.4°
θ = 180° + 70.6° = 250.6°
But 0° ≤ x ≤ 360° → 0° ≤ 3x ≤ 1080° → so find all θ in [0°, 1080°] that satisfy cos θ = -0.3321
Add 360° repeatedly:
First cycle:
- 109.4°, 250.6°
Second cycle (+360):
- 109.4° + 360° = 469.4°
- 250.6° + 360° = 610.6°
Third cycle (+720):
- 109.4° + 720° = 829.4°
- 250.6° + 720° = 970.6°
Now divide each by 3 to get x:
- 109.4° / 3 ≈ 36.5°
- 250.6° / 3 ≈ 83.5°
- 469.4° / 3 ≈ 156.5°
- 610.6° / 3 ≈ 203.5°
- 829.4° / 3 ≈ 276.5°
- 970.6° / 3 ≈ 323.5°
✔ Answers: 36.5°, 83.5°, 156.5°, 203.5°, 276.5°, 323.5°
---
16) sin(x + 20°) = 0.6551
→ Let θ = x + 20° → sin θ = 0.6551 → θ ≈ arcsin(0.6551) ≈ 40.9° (Q1), and 180° - 40.9° = 139.1° (Q2)
So:
- x + 20° = 40.9° → x = 20.9°
- x + 20° = 139.1° → x = 119.1°
Also, since sine repeats every 360°, add 360° to θ:
- θ = 40.9° + 360° = 400.9° → x = 380.9° → too big (>360°)
- θ = 139.1° + 360° = 499.1° → x = 479.1° → too big
So only two solutions.
✔ Answers: 20.9°, 119.1°
---
17) tan(x - 15°) = -0.9128
→ Let θ = x - 15° → tan θ = -0.9128 → reference angle: arctan(0.9128) ≈ 42.4°
Tangent negative in Q2 and Q4:
θ = 180° - 42.4° = 137.6°
θ = 360° - 42.4° = 317.6°
Also, tangent has period 180°, so add 180° to each:
θ = 137.6° + 180° = 317.6° → already have
θ = 317.6° + 180° = 497.6° → check if within range?
Since 0° ≤ x ≤ 360° → θ = x - 15° → -15° ≤ θ ≤ 345°
Wait — let’s be careful.
Actually, θ can go from -15° to 345°, but tangent is periodic with 180°, so we should consider all θ such that tan θ = -0.9128 and θ ∈ [-15°, 345°]
We have:
θ₁ = 137.6° → x = 137.6° + 15° = 152.6°
θ₂ = 317.6° → x = 317.6° + 15° = 332.6°
What about subtracting 180°? θ = 137.6° - 180° = -42.4° → which is ≥ -15°? No, -42.4 < -15 → invalid.
θ = 317.6° - 180° = 137.6° → already have.
So only two solutions.
✔ Answers: 152.6°, 332.6°
---
18) cos(2x + 33°) = 0.306
→ Let θ = 2x + 33° → cos θ = 0.306 → θ ≈ arccos(0.306) ≈ 72.2° (Q1), and 360° - 72.2° = 287.8° (Q4)
Now, 0° ≤ x ≤ 360° → 33° ≤ θ ≤ 2*360° + 33° = 753°
So find all θ in [33°, 753°] where cos θ = 0.306
Start with base solutions:
θ₁ = 72.2°
θ₂ = 287.8°
Add 360°:
θ₃ = 72.2° + 360° = 432.2°
θ₄ = 287.8° + 360° = 647.8°
Add another 360°? 72.2° + 720° = 792.2° > 753° → stop.
So four values: 72.2°, 287.8°, 432.2°, 647.8°
Now solve for x:
θ = 2x + 33° → 2x = θ - 33° → x = (θ - 33°)/2
- x₁ = (72.2 - 33)/2 = 39.2/2 = 19.6°
- x₂ = (287.8 - 33)/2 = 254.8/2 = 127.4°
- x₃ = (432.2 - 33)/2 = 399.2/2 = 199.6°
- x₄ = (647.8 - 33)/2 = 614.8/2 = 307.4°
✔ Answers: 19.6°, 127.4°, 199.6°, 307.4°
---
19) tan²x = ¼
→ Take square root: tan x = ±½
Case 1: tan x = 0.5 → as in #6: x ≈ 26.6° and 206.6°? Wait no — earlier for 0.2 it was 11.3°, now 0.5:
arctan(0.5) ≈ 26.565° → round to 26.6°
So solutions: 26.6°, 180° + 26.6° = 206.6°
Case 2: tan x = -0.5 → reference angle 26.6°, negative in Q2 and Q4:
x = 180° - 26.6° = 153.4°
x = 360° - 26.6° = 333.4°
✔ Answers: 26.6°, 153.4°, 206.6°, 333.4°
---
20) cos²x = ½
→ cos x = ±√(½) = ±√2/2 ≈ ±0.7071
Case 1: cos x = √2/2 → x = 45°, 315°
Case 2: cos x = -√2/2 → x = 135°, 225°
All exact values.
✔ Answers: 45.0°, 135.0°, 225.0°, 315.0°
---
Final Answer:
1) 90.0°
2) 45.0°, 225.0°
3) 0.0°, 360.0°
4) 30.0°, 150.0°
5) 53.1°, 306.9°
6) 11.3°, 191.3°
7) 64.6°, 295.4°
8) 7.1°, 187.1°
9) 41.8°, 138.2°
10) 135.0°, 315.0°
11) 244.2°, 295.8°
12) 174.3°, 354.3°
13) 202.0°, 338.0°
14) 8.4°, 81.6°, 188.4°, 261.6°
15) 36.5°, 83.5°, 156.5°, 203.5°, 276.5°, 323.5°
16) 20.9°, 119.1°
17) 152.6°, 332.6°
18) 19.6°, 127.4°, 199.6°, 307.4°
19) 26.6°, 153.4°, 206.6°, 333.4°
20) 45.0°, 135.0°, 225.0°, 315.0°
Parent Tip: Review the logic above to help your child master the concept of college maths worksheets.
