Answer :
Sure, let's solve each of the given trigonometric equations step-by-step, approximating to the nearest 0.01 radian.
### (a) [tex]$\sin \theta = -0.0149$[/tex]
To find [tex]$\theta$[/tex] satisfying this equation, we look for [tex]$\theta$[/tex] such that the sine of [tex]$\theta$[/tex] equals [tex]$-0.0149$[/tex]. The principal solution for [tex]$\theta$[/tex] can be found using the arcsine function:
[tex]\[ \theta = \arcsin(-0.0149) \approx -0.01 \text{ radians} \][/tex]
Since sine is also negative in the fourth quadrant, we add [tex]$2\pi$[/tex] to bring it into the interval [tex]$[0, 2\pi)$[/tex]:
[tex]\[ \theta = 2\pi + \arcsin(-0.0149) \approx 6.27 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = -0.01, 6.27 \][/tex]
### (b) [tex]$\cos \theta = 0.9256$[/tex]
To find [tex]$\theta$[/tex] such that the cosine of [tex]$\theta$[/tex] equals [tex]$0.9256$[/tex], we use the arccosine function:
[tex]\[ \theta = \arccos(0.9256) \approx 0.39 \text{ radians} \][/tex]
Since cosine is positive in the fourth quadrant, we use the property of the cosine function:
[tex]\[ \theta = 2\pi - \arccos(0.9256) \approx 5.89 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 0.39, 5.89 \][/tex]
### (c) [tex]$\tan \theta = 0.44$[/tex]
To find [tex]$\theta$[/tex] such that the tangent of [tex]$\theta$[/tex] equals [tex]$0.44$[/tex], we use the arctangent function:
[tex]\[ \theta = \arctan(0.44) \approx 0.41 \text{ radians} \][/tex]
Since tangent has a period of [tex]$\pi$[/tex], the other solution within [tex]$[0, 2\pi)$[/tex] is:
[tex]\[ \theta = \pi + \arctan(0.44) \approx 3.55 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 0.41, 3.55 \][/tex]
### (d) [tex]$\cot \theta = -2.771$[/tex]
Since cotangent is the reciprocal of tangent, we first consider [tex]$\theta$[/tex] such that:
[tex]\[ \cot \theta = -2.771 \implies \tan \theta = \frac{1}{-2.771} \approx -0.36 \][/tex]
Using the arctangent function to find the principal angle:
[tex]\[ \theta = \arctan(-0.36) \approx -0.34 \text{ radians} \][/tex]
Because cotangent has a period of [tex]$\pi$[/tex], the possible solutions are adjusted to fit within [tex]$[0, 2\pi)$[/tex]:
[tex]\[ \theta = \arctan(-0.36) + \pi \approx 2.80 \text{ radians} \][/tex]
To get the second solution:
[tex]\[ \theta = \arctan(-0.36) + \pi - \pi = -0.34 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 2.80, -0.34 \][/tex]
### (e) [tex]$\sec \theta = -3.58$[/tex]
Since secant is the reciprocal of cosine, we find [tex]$\theta$[/tex] such that:
[tex]\[ \sec \theta = -3.58 \implies \cos \theta = \frac{1}{-3.58} \approx -0.28 \][/tex]
Using the arccosine function:
[tex]\[ \theta = \arccos(-0.28) \approx 5.00 \text{ radians} \][/tex]
Since cosine is negative in the second quadrant:
[tex]\[ \theta = \pi - \arccos(-0.28) \approx 1.86 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 5.00, 1.86 \][/tex]
### (f) [tex]$\csc \theta = 1.212$[/tex]
Since cosecant is the reciprocal of sine, we find [tex]$\theta$[/tex] such that:
[tex]\[ \csc \theta = 1.212 \implies \sin \theta = \frac{1}{1.212} \approx 0.825 \][/tex]
Using the arcsine function:
[tex]\[ \theta = \arcsin(0.825) \approx 0.97 \text{ radians} \][/tex]
Since sine is positive in the first and second quadrants:
[tex]\[ \theta = 2\pi - \arcsin(0.825) \approx 5.31 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 0.97, 5.31 \][/tex]
Concluding, the solutions are:
- (a) [tex]$\sin \theta = -0.0149$[/tex]: [tex]$\theta = -0.01, 6.27$[/tex]
- (b) [tex]$\cos \theta = 0.9256$[/tex]: [tex]$\theta = 0.39, 5.89$[/tex]
- (c) [tex]$\tan \theta = 0.44$[/tex]: [tex]$\theta = 0.41, 3.55$[/tex]
- (d) [tex]$\cot \theta = -2.771$[/tex]: [tex]$\theta = 2.80, -0.34$[/tex]
- (e) [tex]$\sec \theta = -3.58$[/tex]: [tex]$\theta = 5.00, 1.86$[/tex]
- (f) [tex]$\csc \theta = 1.212$[/tex]: [tex]$\theta = 0.97, 5.31$[/tex]
### (a) [tex]$\sin \theta = -0.0149$[/tex]
To find [tex]$\theta$[/tex] satisfying this equation, we look for [tex]$\theta$[/tex] such that the sine of [tex]$\theta$[/tex] equals [tex]$-0.0149$[/tex]. The principal solution for [tex]$\theta$[/tex] can be found using the arcsine function:
[tex]\[ \theta = \arcsin(-0.0149) \approx -0.01 \text{ radians} \][/tex]
Since sine is also negative in the fourth quadrant, we add [tex]$2\pi$[/tex] to bring it into the interval [tex]$[0, 2\pi)$[/tex]:
[tex]\[ \theta = 2\pi + \arcsin(-0.0149) \approx 6.27 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = -0.01, 6.27 \][/tex]
### (b) [tex]$\cos \theta = 0.9256$[/tex]
To find [tex]$\theta$[/tex] such that the cosine of [tex]$\theta$[/tex] equals [tex]$0.9256$[/tex], we use the arccosine function:
[tex]\[ \theta = \arccos(0.9256) \approx 0.39 \text{ radians} \][/tex]
Since cosine is positive in the fourth quadrant, we use the property of the cosine function:
[tex]\[ \theta = 2\pi - \arccos(0.9256) \approx 5.89 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 0.39, 5.89 \][/tex]
### (c) [tex]$\tan \theta = 0.44$[/tex]
To find [tex]$\theta$[/tex] such that the tangent of [tex]$\theta$[/tex] equals [tex]$0.44$[/tex], we use the arctangent function:
[tex]\[ \theta = \arctan(0.44) \approx 0.41 \text{ radians} \][/tex]
Since tangent has a period of [tex]$\pi$[/tex], the other solution within [tex]$[0, 2\pi)$[/tex] is:
[tex]\[ \theta = \pi + \arctan(0.44) \approx 3.55 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 0.41, 3.55 \][/tex]
### (d) [tex]$\cot \theta = -2.771$[/tex]
Since cotangent is the reciprocal of tangent, we first consider [tex]$\theta$[/tex] such that:
[tex]\[ \cot \theta = -2.771 \implies \tan \theta = \frac{1}{-2.771} \approx -0.36 \][/tex]
Using the arctangent function to find the principal angle:
[tex]\[ \theta = \arctan(-0.36) \approx -0.34 \text{ radians} \][/tex]
Because cotangent has a period of [tex]$\pi$[/tex], the possible solutions are adjusted to fit within [tex]$[0, 2\pi)$[/tex]:
[tex]\[ \theta = \arctan(-0.36) + \pi \approx 2.80 \text{ radians} \][/tex]
To get the second solution:
[tex]\[ \theta = \arctan(-0.36) + \pi - \pi = -0.34 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 2.80, -0.34 \][/tex]
### (e) [tex]$\sec \theta = -3.58$[/tex]
Since secant is the reciprocal of cosine, we find [tex]$\theta$[/tex] such that:
[tex]\[ \sec \theta = -3.58 \implies \cos \theta = \frac{1}{-3.58} \approx -0.28 \][/tex]
Using the arccosine function:
[tex]\[ \theta = \arccos(-0.28) \approx 5.00 \text{ radians} \][/tex]
Since cosine is negative in the second quadrant:
[tex]\[ \theta = \pi - \arccos(-0.28) \approx 1.86 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 5.00, 1.86 \][/tex]
### (f) [tex]$\csc \theta = 1.212$[/tex]
Since cosecant is the reciprocal of sine, we find [tex]$\theta$[/tex] such that:
[tex]\[ \csc \theta = 1.212 \implies \sin \theta = \frac{1}{1.212} \approx 0.825 \][/tex]
Using the arcsine function:
[tex]\[ \theta = \arcsin(0.825) \approx 0.97 \text{ radians} \][/tex]
Since sine is positive in the first and second quadrants:
[tex]\[ \theta = 2\pi - \arcsin(0.825) \approx 5.31 \text{ radians} \][/tex]
Thus, the solutions are:
[tex]\[ \theta = 0.97, 5.31 \][/tex]
Concluding, the solutions are:
- (a) [tex]$\sin \theta = -0.0149$[/tex]: [tex]$\theta = -0.01, 6.27$[/tex]
- (b) [tex]$\cos \theta = 0.9256$[/tex]: [tex]$\theta = 0.39, 5.89$[/tex]
- (c) [tex]$\tan \theta = 0.44$[/tex]: [tex]$\theta = 0.41, 3.55$[/tex]
- (d) [tex]$\cot \theta = -2.771$[/tex]: [tex]$\theta = 2.80, -0.34$[/tex]
- (e) [tex]$\sec \theta = -3.58$[/tex]: [tex]$\theta = 5.00, 1.86$[/tex]
- (f) [tex]$\csc \theta = 1.212$[/tex]: [tex]$\theta = 0.97, 5.31$[/tex]
