SOLUTION: Find all possible solutions: a.) 7cosx = 4 - 2sin^2x b.) 2sec^2x + tan^2x = 3 Thank you.

Algebra ->  Trigonometry-basics -> SOLUTION: Find all possible solutions: a.) 7cosx = 4 - 2sin^2x b.) 2sec^2x + tan^2x = 3 Thank you.       Log On


   

Question 872773: Find all possible solutions:
a.) 7cosx = 4 - 2sin^2x
b.) 2sec^2x + tan^2x = 3
Thank you.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find all possible solutions:
a.) 7cosx = 4 - 2sin^2x
------
7cos(x) = 4 - 2(1-cos^2(x))
7cos(x) = 4 - 2 + 2cos^2(x)
----
2cos^2(x) - 7cos(x) +2 = 0
-----
Use the Quadratic Formula to get:
cos(x) = [7 +- sqrt(49-4*2*2)]/4
cos(x) = [7+-sqrt(33)]/4
cos(x) = 0.3139
x = arccos(0.3139) = +-71.7 degrees
----------------------------------------
b.) 2sec^2x + tan^2x = 3
2(1+tan^2(x)) + tan^2(x) = 3
---------
2 + 3tan^2(x) = 3
tan^2(x) = 1/3
----
tan(x) = 1/sqrt(3)
x = 30 deg or x = 180+30 = 210 degrees
-----------------------------------------
Cheers,
Stan H.
====================