How do you solve $secx+5=2secx+3$ for $0<=x<=2pi$ ?

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Answer 1 · 2018-04-11T08:26:13

$color(blue)(pi/3, (5pi)/3)$

Identity:

$color(red)bb(secx=1/cosx)$

$secx+5=2secx+3$

Collect like terms:

$2secx-secx=5-3$

$secx=2$

Using identity:

$1/cosx=2$

$cosx=1/2$

$x=arccos(cosx)=arccos(1/2)=>x=pi/3$

This is in the I quadrant. The cosine is positive so, we also have an angle in the IV quadrant:

$2pi-pi/3=(5pi)/3$

So our solutions for:

$0 <= x <= 2pi$

$color(blue)(pi/3, (5pi)/3)$

How do you solve secx+5=2secx+3secx+5=2secx+3 for 0<=x<=2pi0<=x<=2pi?