if sin x =0.5 what values satisfy this equation in the range -180<=x<=180
You should know three things here:
1. the fact that sin 30° = 1/2 = 0.5 , (use your calculator to confirm)
2. the shape of the sine curve from -360° to 360°
https://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html
3. the CAST rule, which would tell you that the sine function is positive
in quadrants I and II
so x = 30° or x = 180°-30° or 150°
or x = -210° or -330° , look at the yellow graph
oops, forget about -210 and -330,
even though they are solutions, they are not within your given domain
so x = 30° or 150° for your domain
To find the values of x that satisfy the equation sin x = 0.5 within the given range -180 <= x <= 180, we can use the inverse sine function or arcsin.
The inverse sine function is denoted as arcsin or sin^(-1).
Step-by-step solution:
1. Start with the equation sin x = 0.5.
2. Take the inverse sine (arcsin) of both sides to isolate x: arcsin(sin x) = arcsin(0.5).
3. The arcsin and sin functions cancel each other out, leaving x = arcsin(0.5).
4. Use a scientific calculator to find the arcsin (or inverse sine) of 0.5. The result is approximately 30 degrees (or π/6 radians).
5. Since the range given is -180 <= x <= 180, we need to check if there are any other values in this range that satisfy the equation.
6. We know that sin x = 0.5 has two solutions, one positive and one negative, symmetric about the y-axis. The positive solution is x = 30 degrees (or π/6 radians).
7. To find the negative solution, we subtract the positive solution from 180 degrees (or π radians): x_negative = 180 - 30 = 150 degrees (or 5π/6 radians).
8. Therefore, the values of x that satisfy the equation sin x = 0.5 in the range -180 <= x <= 180 are x = 30 degrees and x = 150 degrees (or π/6 radians and 5π/6 radians).