cos²(θ) – 3sin(θ) – sin²(θ)= –2
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I am having a hard time figuring out what to do. Do I use the Pythagorean ID: sin²(θ) + cos²(θ) first?
2 answers
Sorry, I also forgot: I have to find all the solutions based on the interval [0 deg., 360 deg.)
yes ... cos^2(Θ) = 1 - sin^2(Θ)
1 - sin^2(Θ) - 3 sin(Θ) - sin^2(Θ) = -2
2 sin^2(Θ) + 3 sin(Θ) - 3 = 0 ... solve the quadratic for sin(Θ)
... or factor
1 - sin^2(Θ) - 3 sin(Θ) - sin^2(Θ) = -2
2 sin^2(Θ) + 3 sin(Θ) - 3 = 0 ... solve the quadratic for sin(Θ)
... or factor
3 answers