Math  /  Trigonometry

QuestionFor the following exercises, solve exactly on [0,2π)[0,2 \pi).
13. 2cosθ=22 \cos \theta=\sqrt{2}
14. 2cosθ=12 \cos \theta=-1
15. 2sinθ=12 \sin \theta=-1
16. 2sinθ=32 \sin \theta=-\sqrt{3}
17. 2sin(3θ)=12 \sin (3 \theta)=1
18. 2sin(2θ)=32 \sin (2 \theta)=\sqrt{3}
19. 2cos(3θ)=22 \cos (3 \theta)=-\sqrt{2}
20. cos(2θ)=32\cos (2 \theta)=-\frac{\sqrt{3}}{2}
21. 2sin(πθ)=12 \sin (\pi \theta)=1
22. 2cos(π5θ)=32 \cos \left(\frac{\pi}{5} \theta\right)=\sqrt{3}

For the following exercises, find all exact solutions on [0,2π)[0,2 \pi).

Studdy Solution
Find specific solutions within the interval [0,2π)[0, 2\pi) by substituting integer values for kk.
For k=0k = 0: θ=π18,5π18 \theta = \frac{\pi}{18}, \frac{5\pi}{18}
For k=1k = 1: θ=π18+2π3=13π18 \theta = \frac{\pi}{18} + \frac{2\pi}{3} = \frac{13\pi}{18} θ=5π18+2π3=17π18 \theta = \frac{5\pi}{18} + \frac{2\pi}{3} = \frac{17\pi}{18}
For k=2k = 2: θ=π18+4π3=25π18 \theta = \frac{\pi}{18} + \frac{4\pi}{3} = \frac{25\pi}{18} θ=5π18+4π3=29π18 \theta = \frac{5\pi}{18} + \frac{4\pi}{3} = \frac{29\pi}{18}
Check if these are within [0,2π)[0, 2\pi). If not, discard them.
The solutions for Problem 17 are:
θ=π18,5π18,13π18,17π18 \theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}
**Problem 18: 2sin(2θ)=32 \sin (2 \theta) = \sqrt{3}**
STEP_1: Isolate sin(2θ)\sin (2 \theta) by dividing both sides by 2:
sin(2θ)=32 \sin (2 \theta) = \frac{\sqrt{3}}{2}
STEP_2: Determine the angles where sin(2θ)=32\sin (2 \theta) = \frac{\sqrt{3}}{2}. These angles are:
2θ=π3,2π3 2\theta = \frac{\pi}{3}, \frac{2\pi}{3}
STEP_3: Since the period of sin\sin is 2π2\pi, we add multiples of 2π2\pi to find all solutions for 2θ2\theta:
2θ=π3+2kπ,2π3+2kπ 2\theta = \frac{\pi}{3} + 2k\pi, \frac{2\pi}{3} + 2k\pi
where kk is an integer.
STEP_4: Solve for θ\theta by dividing each term by 2:
θ=π6+kπ,π3+kπ \theta = \frac{\pi}{6} + k\pi, \frac{\pi}{3} + k\pi
STEP_5: Find specific solutions within the interval [0,2π)[0, 2\pi) by substituting integer values for kk.
For k=0k = 0: θ=π6,π3 \theta = \frac{\pi}{6}, \frac{\pi}{3}
For k=1k = 1: θ=π6+π=7π6 \theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6} θ=π3+π=4π3 \theta = \frac{\pi}{3} + \pi = \frac{4\pi}{3}
Check if these are within [0,2π)[0, 2\pi). If not, discard them.
The solutions for Problem 18 are:
θ=π6,π3,7π6,4π3 \theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}
**Problem 19: 2cos(3θ)=22 \cos (3 \theta) = -\sqrt{2}**
STEP_1: Isolate cos(3θ)\cos (3 \theta) by dividing both sides by 2:
cos(3θ)=22 \cos (3 \theta) = -\frac{\sqrt{2}}{2}
STEP_2: Determine the angles where cos(3θ)=22\cos (3 \theta) = -\frac{\sqrt{2}}{2}. These angles are:
3θ=3π4,5π4 3\theta = \frac{3\pi}{4}, \frac{5\pi}{4}
STEP_3: Since the period of cos\cos is 2π2\pi, we add multiples of 2π2\pi to find all solutions for 3θ3\theta:
3θ=3π4+2kπ,5π4+2kπ 3\theta = \frac{3\pi}{4} + 2k\pi, \frac{5\pi}{4} + 2k\pi
where kk is an integer.
STEP_4: Solve for θ\theta by dividing each term by 3:
θ=π4+2kπ3,5π12+2kπ3 \theta = \frac{\pi}{4} + \frac{2k\pi}{3}, \frac{5\pi}{12} + \frac{2k\pi}{3}
STEP_5: Find specific solutions within the interval [0,2π)[0, 2\pi) by substituting integer values for kk.
For k=0k = 0: θ=π4,5π12 \theta = \frac{\pi}{4}, \frac{5\pi}{12}
For k=1k = 1: θ=π4+2π3=11π12 \theta = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{11\pi}{12} θ=5π12+2π3=13π12 \theta = \frac{5\pi}{12} + \frac{2\pi}{3} = \frac{13\pi}{12}
For k=2k = 2: θ=π4+4π3=19π12 \theta = \frac{\pi}{4} + \frac{4\pi}{3} = \frac{19\pi}{12} θ=5π12+4π3=23π12 \theta = \frac{5\pi}{12} + \frac{4\pi}{3} = \frac{23\pi}{12}
Check if these are within [0,2π)[0, 2\pi). If not, discard them.
The solutions for Problem 19 are:
θ=π4,5π12,11π12,13π12,19π12,23π12 \theta = \frac{\pi}{4}, \frac{5\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}
**Problem 20: cos(2θ)=32\cos (2 \theta) = -\frac{\sqrt{3}}{2}**
STEP_1: Isolate cos(2θ)\cos (2 \theta):
cos(2θ)=32 \cos (2 \theta) = -\frac{\sqrt{3}}{2}
STEP_2: Determine the angles where cos(2θ)=32\cos (2 \theta) = -\frac{\sqrt{3}}{2}. These angles are:
2θ=5π6,7π6 2\theta = \frac{5\pi}{6}, \frac{7\pi}{6}
STEP_3: Since the period of cos\cos is 2π2\pi, we add multiples of 2π2\pi to find all solutions for 2θ2\theta:
2θ=5π6+2kπ,7π6+2kπ 2\theta = \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k\pi
where kk is an integer.
STEP_4: Solve for θ\theta by dividing each term by 2:
θ=5π12+kπ,7π12+kπ \theta = \frac{5\pi}{12} + k\pi, \frac{7\pi}{12} + k\pi
STEP_5: Find specific solutions within the interval [0,2π)[0, 2\pi) by substituting integer values for kk.
For k=0k = 0: θ=5π12,7π12 \theta = \frac{5\pi}{12}, \frac{7\pi}{12}
For k=1k = 1: θ=5π12+π=17π12 \theta = \frac{5\pi}{12} + \pi = \frac{17\pi}{12} θ=7π12+π=19π12 \theta = \frac{7\pi}{12} + \pi = \frac{19\pi}{12}
Check if these are within [0,2π)[0, 2\pi). If not, discard them.
The solutions for Problem 20 are:
θ=5π12,7π12,17π12,19π12 \theta = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}
**Problem 21: 2sin(πθ)=12 \sin (\pi \theta) = 1**
STEP_1: Isolate sin(πθ)\sin (\pi \theta) by dividing both sides by 2:
sin(πθ)=12 \sin (\pi \theta) = \frac{1}{2}
STEP_2: Determine the angles where sin(πθ)=12\sin (\pi \theta) = \frac{1}{2}. These angles are:
πθ=π6,5π6 \pi \theta = \frac{\pi}{6}, \frac{5\pi}{6}
STEP_3: Since the period of sin\sin is 2π2\pi, we add multiples of 2π2\pi to find all solutions for πθ\pi \theta:
πθ=π6+2kπ,5π6+2kπ \pi \theta = \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi
where kk is an integer.
STEP_4: Solve for θ\theta by dividing each term by π\pi:
θ=16+2k,56+2k \theta = \frac{1}{6} + 2k, \frac{5}{6} + 2k
STEP_5: Find specific solutions within the interval [0,2π)[0, 2\pi) by substituting integer values for kk.
For k=0k = 0: θ=16,56 \theta = \frac{1}{6}, \frac{5}{6}
For k=1k = 1: θ=16+2=136 \theta = \frac{1}{6} + 2 = \frac{13}{6} θ=56+2=176 \theta = \frac{5}{6} + 2 = \frac{17}{6}
Check if these are within [0,2π)[0, 2\pi). If not, discard them.
The solutions for Problem 21 are:
θ=16,56,136,176 \theta = \frac{1}{6}, \frac{5}{6}, \frac{13}{6}, \frac{17}{6}
**Problem 22: 2cos(π5θ)=32 \cos \left(\frac{\pi}{5} \theta\right) = \sqrt{3}**
STEP_1: Isolate cos(π5θ)\cos \left(\frac{\pi}{5} \theta\right) by dividing both sides by 2:
cos(π5θ)=32 \cos \left(\frac{\pi}{5} \theta\right) = \frac{\sqrt{3}}{2}
STEP_

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