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?
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
Sorry, I also forgot: I have to find all the solutions based on the interval [0 deg., 360 deg.)
Yes, you're on the right track. To solve the given equation, you can indeed utilize the Pythagorean identity: sin²(θ) + cos²(θ) = 1.
Let's rearrange the original equation by moving the terms to one side to obtain a quadratic equation:
cos²(θ) - sin²(θ) - 3sin(θ) = -2
Now, substitute sin²(θ) with 1 - cos²(θ) using the Pythagorean identity:
cos²(θ) - (1 - cos²(θ)) - 3sin(θ) = -2
Next, distribute the negative sign:
cos²(θ) - 1 + cos²(θ) - 3sin(θ) = -2
Combine like terms:
2cos²(θ) - 3sin(θ) - 1 = -2
Rearrange and simplify:
2cos²(θ) - 3sin(θ) + 1 = 0
Now you have a quadratic equation in terms of cos(θ) and sin(θ). To solve it, you can use the quadratic formula or factorization techniques. However, before you proceed, consider substituting sin(θ) with √(1 - cos²(θ)) to eliminate the trigonometric function sin(θ) from the equation.
By substituting sin(θ) with √(1 - cos²(θ)), you will obtain a quadratic equation solely in terms of cos(θ):
2cos²(θ) - 3√(1 - cos²(θ)) + 1 = 0
Now you can apply algebraic techniques to solve for cos(θ), such as factoring, completing the square, or using the quadratic formula.
Once you find the value(s) of cos(θ), you can substitute it back into the original equation sin²(θ) + cos²(θ) = 1 to solve for sin(θ), if necessary.