Question

How do you solve the equation on the interval?

`[0,2pi)`

`15 sec^2 x-20=0`

Answer 1

`x=pi/6, (5pi)/6, (7pi)/6, (11pi)/6`

Explanation:

Given:
`15sec^2x-20=0`
Transposing 20 to rhs
`15sec^2x=20`
Dividing by 15
`sec^2x=20/15=4/3`
`sec^2x=1/cos^2x`
`4/3=1/cos^2x`
`cos^2x=3/4`
`cosx=+-sqrt(3/4)`
`cosx=+-sqrt3/2`
cosx is positive in first and fourth quadrant
`cos(pi/6)=sqrt3/2`
`cos(2pi-pi/6)-sqrt3/2`
cosx is negative in second and third quadrant
`cos(pi-pi/6)=-sqrt3/2`
`cos(pi+pi/6)=-sqrt3/2`
The values of x lying within the interval
`(0,2pi)are, x=pi/6,2pi-pi/6,pi-pi/6,pi+pi/6`
when arranged in increasing order
`x=pi/6, (5pi)/6, (7pi)/6, (11pi)/6`