Solve the equation on the interval 0≤θ<2π sin^2 θ + 17 sin θ +9=0
Added by Lu-S T.
Step 1
So, we can use the quadratic formula to solve for sin θ: sin θ = (-b ± √(b^2 - 4ac)) / 2a where a = 1, b = 17, and c = 9. Plugging these values in, we get: sin θ = (-17 ± √(17^2 - 4(1)(9))) / 2(1) sin θ = (-17 ± √265) / 2 Now, we need to check if these Show more…
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