Solve each equation to the nearest tenth and use the given restrictions.
sin θ = -0.204 for 90 degrees < θ < 270 degrees.
I don't get this. Is this supposed to be restricted to quadrants 2 and 3? Then it doesn't seem possible to get a value that will be equal to -0.204.
yes. Since sinθ is negative in QIII and QIV, that should make it clear that they only want one solution.
sin(.2054) = .204, you want QIII, so θ=π+.2054
You are correct in your understanding of the restrictions. The equation sin θ = -0.204, with the given restriction 90 degrees < θ < 270 degrees, means that we are looking for an angle θ within that range where the sine of that angle is equal to -0.204.
In the standard unit circle, sine is positive in the second and third quadrants, and negative in the third and fourth quadrants. Therefore, we can conclude that θ lies in either the second or third quadrant.
To solve the equation, we can use the inverse sine function (sin^(-1)) to find the angle whose sine is equal to -0.204:
θ = sin^(-1)(-0.204)
Using a scientific calculator, you can find the inverse sine of -0.204. The result will be an angle measured in radians. Next, convert the radians to degrees.
θ ≈ -11.76 degrees
However, this value is outside the given restriction of 90 degrees < θ < 270 degrees. Therefore, there is no solution within this range that satisfies the equation sin θ = -0.204 with the given restrictions.