Thread: Trig problem, ill paypal you 5 dollars if its right.

1. Need help, dont know how to solve, if you do, ill paypal you 5 dollars if its right.

cos x csc^2 x + 3 cos x = 7 cos x

Thanks,
Mike

2. cos x csc^2 x + 3 cos x = 7 cos x

Rewriting that,
cosXcsc^2(X) +3cosX = 7cosX
cosXcsc^2(X) +3cosX -7cosX = 0
cosXcsc^2(X) -4cosX = 0
cosX[csc^2(X) -4] = 0

cosX = 0
X = arccos(0) = 90 or 270 degrees

csc^2(X) -4 = 0
csc^2(X) = 4
cscX = +,-2

When cscX = 2,
cscX = 1/sinX
So, 1/sinX = 2
sinX = 1/2 ---positive
Sine is positive in the 1st and 2nd quadrants, so,
X = arcsin(1/2)
X = 30 deg --in the 1st quadrant
X = 180-30 = 150 deg ---in the 2nd quarter

When cscX = -2,
1/sinX = -2
sinX = -1/2 ---negative
Sine is negative in the 3rd and 4th quadrants, so,
X = arcsin(-1/2)
X = 180 +30 = 210 deg --in the 3rd quadrant
X = 360 -30 = 330 deg ---in the 4th quarter

Therefore, if X is from 0 to 360 degrees only,
X = 30, 90, 150, 210, 270, or 330 degrees -------answer.

Check those against the original equation and you'd find them all correct.

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