SOLUTION: 7 tan alpha + cot alpha = 5 sec alpha

Question 1088729: 7 tan alpha + cot alpha = 5 sec alpha

Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
7 tan alpha + cot alpha = 5 sec alpha
Multiply thru by cosine
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7sin + cos^2/sin = 5
7sin^2 + cos^2 = 5sin
6sin^2 + sin^2 + cos^2 = 5sin
6sin^2 + 1 = 5sin
6sin^2 - 5sin + 1 = 0
(3sin - 1)*(2sin - 1) = 0
etc.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
7(sin x/cos x)+cos x/sin x=5/cos x
(1/sin x*cos x)(7 sin^2 x+cos^2 x)=5/cos x; common denominator
(1/ sin x)(7 sin^2 x+ cos^2 x)=5; multiplying both sides by cos x get the next line
7 sin^2 x+ cos^2 x=5 sin x
7 sin^2x-5 sin x+1-sin^2x=0, using identity for cos^2x=1-sin^2x
6 sin^2x-5 sin x+1=0; collecting terms
(3 sin x-1)( 2sin x-1)=0;factoring
sin x=1/3, that occurs at 19.47 and 160.53 degrees
sin x=1/2; that occurs at 30 and 150 degrees
check with 30 degrees.
7 tan x=7*sqrt(3)/3
ctn x=3/sqrt(3)=3*sqrt(3)/3
5/cos x=5/(sqrt(3)/2=10/sqrt(3)=10 sqrt(3)/3. The first two add to the third.
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at 150 degrees, the tangent is negative, as is the cotangent. They will sum to the same negative and the cosine is also negative in the second quadrant, so that checks.
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7 tangent 19.47 deg=2.47
ctn 19.47=2.83
5/cos 19.47=5.30, and that checks
In the second quadrant, all are negative.
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