Join the IDNLearn.com community and start finding the answers you need today. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.
Sagot :
To find the angles between \(0^\circ\) and \(360^\circ\) (inclusive) that satisfy the given equations, we will go through each equation step-by-step.
### Part a) \(2 \sin (x + 50^\circ) = 1\)
1. Isolate the sine term:
[tex]\[ 2 \sin (x + 50^\circ) = 1 \][/tex]
[tex]\[ \sin (x + 50^\circ) = \frac{1}{2} \][/tex]
2. Find the general solutions for the sine function:
Since \(\sin \theta = \frac{1}{2}\) at \(\theta = 30^\circ + 360^\circ n\) and \(150^\circ + 360^\circ n\) for integer values of \(n\).
3. Adjust for the phase shift:
[tex]\[ x + 50^\circ = 30^\circ + 360^\circ n \][/tex]
[tex]\[ x + 50^\circ = 150^\circ + 360^\circ n \][/tex]
4. Solve for \(x\):
[tex]\[ x = 30^\circ - 50^\circ + 360^\circ n = -20^\circ + 360^\circ n \][/tex]
[tex]\[ x = 150^\circ - 50^\circ + 360^\circ n = 100^\circ + 360^\circ n \][/tex]
[tex]\[ x = 30^\circ + 360^\circ n - 50^\circ = -20^\circ + 360^\circ n \][/tex]
[tex]\[ x = 150^\circ + 360^\circ n - 50^\circ = 100^\circ + 360^\circ n \][/tex]
5. Plug in values of \(n\) to find solutions within \(0^\circ\) to \(360^\circ\):
For \(n = 0\):
[tex]\[ x = -20^\circ \][/tex] (Not in the range 0° to 360°)
[tex]\[ x = 100^\circ \][/tex]
For \(n = 1\):
[tex]\[ x = 340^\circ \][/tex]
Therefore, the solutions are \(100^\circ\) and \(340^\circ\).
### Part b) \(\cos 3x = \sin \frac{\pi}{6}\)
1. Recognize that \(\sin \frac{\pi}{6} = \frac{1}{2}\):
[tex]\[ \cos 3x = \frac{1}{2} \][/tex]
2. Find the general solutions for the cosine function:
Since \(\cos \theta = \frac{1}{2}\) at \(\theta = 60^\circ + 360^\circ n\) and \(300^\circ + 360^\circ n\) for integer values of \(n\).
3. Adjust for the coefficient \(3x\):
[tex]\[ 3x = 60^\circ + 360^\circ n \][/tex]
[tex]\[ 3x = 300^\circ + 360^\circ n \][/tex]
4. Solve for \(x\):
[tex]\[ x = \frac{60^\circ + 360^\circ n}{3} = 20^\circ + 120^\circ n \][/tex]
[tex]\[ x = \frac{300^\circ + 360^\circ n}{3} = 100^\circ + 120^\circ n \][/tex]
5. Plug in values of \(n\) to find solutions within \(0^\circ\) to \(360^\circ\):
For \(n = 0\),
[tex]\[ x = 20^\circ \][/tex]
[tex]\[ x = 100^\circ \][/tex]
For \(n = 1\),
[tex]\[ x = 140^\circ \][/tex]
[tex]\[ x = 220^\circ \][/tex]
For \(n = 2\),
[tex]\[ x = 260^\circ \][/tex]
[tex]\[ x = 340^\circ \][/tex]
Therefore, the solutions are \(20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, \) and \(340^\circ\).
### Part c) \(\cos (2x + 20^\circ) = -\frac{1}{2}\)
1. Isolate the cosine term:
[tex]\[ \cos (2x + 20^\circ) = -\frac{1}{2} \][/tex]
2. Find the general solutions for the cosine function:
Since \(\cos \theta = -\frac{1}{2}\) at \(\theta = 120^\circ + 360^\circ n\) and \(240^\circ + 360^\circ n\) for integer values of \(n\).
3. Adjust for the phase shift:
[tex]\[ 2x + 20^\circ = 120^\circ + 360^\circ n \][/tex]
[tex]\[ 2x + 20^\circ = 240^\circ + 360^\circ n \][/tex]
4. Solve for \(x\):
[tex]\[ 2x = 120^\circ - 20^\circ + 360^\circ n = 100^\circ + 360^\circ n \][/tex]
[tex]\[ 2x = 240^\circ - 20^\circ + 360^\circ n = 220^\circ + 360^\circ n \][/tex]
[tex]\[ x = \frac{100^\circ + 360^\circ n}{2} = 50^\circ + 180^\circ n \][/tex]
[tex]\[ x = \frac{220^\circ + 360^\circ n}{2} = 110^\circ + 180^\circ n \][/tex]
5. Plug in values of \(n\) to find solutions within \(0^\circ\) to \(360^\circ\):
For \(n = 0\),
[tex]\[ x = 50^\circ \][/tex]
[tex]\[ x = 110^\circ \][/tex]
For \(n = 1\),
[tex]\[ x = 230^\circ \][/tex]
[tex]\[ x = 290^\circ \][/tex]
Therefore, the solutions are \(50^\circ, 110^\circ, 230^\circ,\) and \(290^\circ\).
### Summary
- For part a), the angles that satisfy the equation are: \(100^\circ\) and \(340^\circ\).
- For part b), the angles that satisfy the equation are: \(20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ,\) and \(340^\circ\).
- For part c), the angles that satisfy the equation are: [tex]\(50^\circ, 110^\circ, 230^\circ,\)[/tex] and [tex]\(290^\circ\)[/tex].
### Part a) \(2 \sin (x + 50^\circ) = 1\)
1. Isolate the sine term:
[tex]\[ 2 \sin (x + 50^\circ) = 1 \][/tex]
[tex]\[ \sin (x + 50^\circ) = \frac{1}{2} \][/tex]
2. Find the general solutions for the sine function:
Since \(\sin \theta = \frac{1}{2}\) at \(\theta = 30^\circ + 360^\circ n\) and \(150^\circ + 360^\circ n\) for integer values of \(n\).
3. Adjust for the phase shift:
[tex]\[ x + 50^\circ = 30^\circ + 360^\circ n \][/tex]
[tex]\[ x + 50^\circ = 150^\circ + 360^\circ n \][/tex]
4. Solve for \(x\):
[tex]\[ x = 30^\circ - 50^\circ + 360^\circ n = -20^\circ + 360^\circ n \][/tex]
[tex]\[ x = 150^\circ - 50^\circ + 360^\circ n = 100^\circ + 360^\circ n \][/tex]
[tex]\[ x = 30^\circ + 360^\circ n - 50^\circ = -20^\circ + 360^\circ n \][/tex]
[tex]\[ x = 150^\circ + 360^\circ n - 50^\circ = 100^\circ + 360^\circ n \][/tex]
5. Plug in values of \(n\) to find solutions within \(0^\circ\) to \(360^\circ\):
For \(n = 0\):
[tex]\[ x = -20^\circ \][/tex] (Not in the range 0° to 360°)
[tex]\[ x = 100^\circ \][/tex]
For \(n = 1\):
[tex]\[ x = 340^\circ \][/tex]
Therefore, the solutions are \(100^\circ\) and \(340^\circ\).
### Part b) \(\cos 3x = \sin \frac{\pi}{6}\)
1. Recognize that \(\sin \frac{\pi}{6} = \frac{1}{2}\):
[tex]\[ \cos 3x = \frac{1}{2} \][/tex]
2. Find the general solutions for the cosine function:
Since \(\cos \theta = \frac{1}{2}\) at \(\theta = 60^\circ + 360^\circ n\) and \(300^\circ + 360^\circ n\) for integer values of \(n\).
3. Adjust for the coefficient \(3x\):
[tex]\[ 3x = 60^\circ + 360^\circ n \][/tex]
[tex]\[ 3x = 300^\circ + 360^\circ n \][/tex]
4. Solve for \(x\):
[tex]\[ x = \frac{60^\circ + 360^\circ n}{3} = 20^\circ + 120^\circ n \][/tex]
[tex]\[ x = \frac{300^\circ + 360^\circ n}{3} = 100^\circ + 120^\circ n \][/tex]
5. Plug in values of \(n\) to find solutions within \(0^\circ\) to \(360^\circ\):
For \(n = 0\),
[tex]\[ x = 20^\circ \][/tex]
[tex]\[ x = 100^\circ \][/tex]
For \(n = 1\),
[tex]\[ x = 140^\circ \][/tex]
[tex]\[ x = 220^\circ \][/tex]
For \(n = 2\),
[tex]\[ x = 260^\circ \][/tex]
[tex]\[ x = 340^\circ \][/tex]
Therefore, the solutions are \(20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, \) and \(340^\circ\).
### Part c) \(\cos (2x + 20^\circ) = -\frac{1}{2}\)
1. Isolate the cosine term:
[tex]\[ \cos (2x + 20^\circ) = -\frac{1}{2} \][/tex]
2. Find the general solutions for the cosine function:
Since \(\cos \theta = -\frac{1}{2}\) at \(\theta = 120^\circ + 360^\circ n\) and \(240^\circ + 360^\circ n\) for integer values of \(n\).
3. Adjust for the phase shift:
[tex]\[ 2x + 20^\circ = 120^\circ + 360^\circ n \][/tex]
[tex]\[ 2x + 20^\circ = 240^\circ + 360^\circ n \][/tex]
4. Solve for \(x\):
[tex]\[ 2x = 120^\circ - 20^\circ + 360^\circ n = 100^\circ + 360^\circ n \][/tex]
[tex]\[ 2x = 240^\circ - 20^\circ + 360^\circ n = 220^\circ + 360^\circ n \][/tex]
[tex]\[ x = \frac{100^\circ + 360^\circ n}{2} = 50^\circ + 180^\circ n \][/tex]
[tex]\[ x = \frac{220^\circ + 360^\circ n}{2} = 110^\circ + 180^\circ n \][/tex]
5. Plug in values of \(n\) to find solutions within \(0^\circ\) to \(360^\circ\):
For \(n = 0\),
[tex]\[ x = 50^\circ \][/tex]
[tex]\[ x = 110^\circ \][/tex]
For \(n = 1\),
[tex]\[ x = 230^\circ \][/tex]
[tex]\[ x = 290^\circ \][/tex]
Therefore, the solutions are \(50^\circ, 110^\circ, 230^\circ,\) and \(290^\circ\).
### Summary
- For part a), the angles that satisfy the equation are: \(100^\circ\) and \(340^\circ\).
- For part b), the angles that satisfy the equation are: \(20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ,\) and \(340^\circ\).
- For part c), the angles that satisfy the equation are: [tex]\(50^\circ, 110^\circ, 230^\circ,\)[/tex] and [tex]\(290^\circ\)[/tex].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.