Answer :
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].
