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Trig question
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5 cos x - 12 sin x = -3 find x I'll post my working... please let me finish first
So far I have: -12 sin x + 5 cos x = -3 13 sin (x - 22.62) = -3 sin (x - 22.62) = -3/13 x - 22.62 = -13.34 x = 9.28 and 170.72 These don't work though
I think when you arc sin both sides you had your calculator in degree mode instead of radians.
@Romero, sorry I forgot to put 0° ≤ x ≤ 360°
\[5 \cos x - 12 \sin x = -3\]\[5 \cos x = 12 \sin x - 3\]square both sides\[25 \ \cos^{2} x = 144 \ \sin^{2} x - 72 \sin x + 9\]\[25(1 - \sin^2 x) = 144 \ \sin^2 x - 72 \sin x + 9\]\[169 \ \sin^2 x - 72 \sin x - 16 = 0\]use the quadratic formula to solve for sin x
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Your answer should involve either Arccos or ArcSin.
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