How do you solve 1+secx=2cosx1+secx=2cosx?

1 Answer
Dec 30, 2016

0, (2pi)/3, (4pi)/3, 2pi0,2π3,4π3,2π

Explanation:

1 + sec x = 2cos x
1 + 1/(cos x) = 2cos x1+1cosx=2cosx
cos x + 1 = 2cos^2 xcosx+1=2cos2x
2cos^2 x - cos x - 1 = 02cos2xcosx1=0
Solve this quadratic equation for cos x.
Use trig table and unit circle.
Since a + b + c = 0, use shortcut. There are 2 real roots:
cos x = 1 and cos x = c/a = - 1/2.
a. cos x = 1 --> x = 0 and x = 2pix=2π
b. cos x = - 1/2cosx=12 -->arc x = +- (2pi)/3x=±2π3
Co-terminal of arc (-2pi)/32π3 is arc (4pi)/34π3
Answers for (0, 2pi)(0,2π):
0, (2pi)/3, (4pi)/3, 2pi0,2π3,4π3,2π
For general answers add 2kpi2kπ