SOLUTION: Find all solutions in the interval [0, 2π) {{{ sec^2(x)+6tan(x)=6 }}} Thank you friends

Algebra ->  Trigonometry-basics -> SOLUTION: Find all solutions in the interval [0, 2π) {{{ sec^2(x)+6tan(x)=6 }}} Thank you friends      Log On


   

Question 1101279: Find all solutions in the interval [0, 2π)
+sec%5E2%28x%29%2B6tan%28x%29=6+
Thank you friends
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!

+sec%5E2%28x%29%2B6tan%28x%29=6+
------
1+%2B+tan%5E2%28x%29%2B6tan%28x%29=6+
tan%5E2%28x%29+%2B+6tan%28x%29+-+5+=+0
tan(x) = -3 + sqrt(14)
tan(x) = -3 - sqrt(14)
etc

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
+sec%5E2%28x%29%2B6tan%28x%29=6+

%281%2Btan%5E2%28x%29%29%2B6tan%28x%29=6+

tan%5E2%28x%29%2B6tan%28x%29%2B1=6+

tan%5E2%28x%29%2B6tan%28x%29-5=0+

tan%28x%29%2B6tan%28x%29-5=0+

tan%28x%29+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A%28-5%29+%29%29%2F%282%2A1%29+

tan%28x%29+=+%28-6+%2B-+sqrt%2836%2B20+%29%29%2F2+

tan%28x%29+=+%28-6+%2B-+sqrt%2856%29%29%2F2+

tan%28x%29+=+%28-6+%2B-+sqrt%284%2A14%29%29%2F2+

tan%28x%29+=+%28-6+%2B-+2sqrt%2814%29%29%2F2+

tan%28x%29+=+%28-6%29%2F2+%2B-+%282sqrt%2814%29%29%2F2+

tan%28x%29+=+-3+%2B-+sqrt%2814%29+

Using the + sign:

tan%28x%29+=+-3+%2B+sqrt%2814%29+

Approximation:

tan%28x%29+=+0.7416573868

The tangent is positive in quadrants I and III.

We find the reference angle by using the absolute value,
that is, we find the inverse tangent of +0.7416573868. 
Make sure your calculator is in radian mode:



To get the answer in quadrant I, use the reference
angle.

x=0.6381404208   <-- that's one solution

To get the answer in quadrant III, add p to the
reference angle.

x=0.6381404208%2B3.141592654

x=3.779733074   <-- that's a second solution

Using the - sign:

tan%28x%29+=+-3+-+sqrt%2814%29+

Approximation:

tan%28x%29+=+-6.741657387

The tangent is negative in quadrants II and IV.

We find the reference angle by using the absolute value,
that is, we find the inverse tangent of +6.741657387. 
Make sure your calculator is still in radian mode:



To get the answer in quadrant II, subtract the reference
angle from p

3.141592654-1.423538584

x=1.71805407   <-- that's a third solution

To get the answer in quadrant IV, subtract the reference
angle from 2p.

x=2%283.141592654%29-1.71805407

x=4.565131237   <-- that's a fourth solution

Edwin