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
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.
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