solve for the equation x
sin x= -.35
180<x<270
sin x= .2
90<x<180
cos x = -.72
180< x<270
please explain, thanks :)
Here is a standard way to solve these kind of problems, I will do the first one
sinx = -.35 , 180 < x <270
1. Find the acute angle, sometimes called the "angle in standard position"
by using the positive decimal
if sinx = .35, x = 20.5°
but x is supposed to be in III , then
x = 180 + 20.5 = 200.5°
check: on calculator, sin 200.5 = -.3502..
try the others the same way, check the answer with your calculator.
Sure! I'd be happy to explain how to solve each of these equations.
1. To solve the equation sin x = -0.35, where 180 < x < 270:
- First, find the reference angle by using the inverse sine function: arcsin(0.35) ≈ 20.95 degrees.
- Since the sine is negative in the specified range (180 < x < 270), subtract the reference angle from 180:
x = 180 - 20.95 ≈ 159.05 degrees.
2. To solve the equation sin x = 0.2, where 90 < x < 180:
- Again, find the reference angle by using the inverse sine function: arcsin(0.2) ≈ 11.53 degrees.
- In this case, the range is 90 < x < 180, so the answer is the reference angle itself:
x ≈ 11.53 degrees.
3. To solve the equation cos x = -0.72, where 180 < x < 270:
- Determine the reference angle using the inverse cosine function: arccos(0.72) ≈ 43.62 degrees.
- Because the cosine is negative in the specified range (180 < x < 270), subtract the reference angle from 360:
x = 360 - 43.62 ≈ 316.38 degrees.
Please note that these solutions are given in degrees. Additionally, remember to always use the appropriate units (degrees or radians) when working with trigonometric functions.