Question

Question #21463

Answer 1

For `0<=theta<2pi`, `theta = pi/5, (5pi)/6, (3pi)/2`

Explanation:

`cos^2theta-sin^2theta=sintheta`

Use the trig identity `sin^2theta + cos^2theta =1`
`=> cos^2theta=1-sin^2theta`

Substitute for `cos^2theta`

`1-sin^2theta-sin^2theta=sintheta`

`-2sin^2theta +1=sintheta`

`0=2sin^2theta+sintheta-1`

Factor

`(2sintheta-1)(sintheta+1)=0`

Set each factor equal to zero and solve.

`2sintheta - 1 =0 color(white)(aaa)sintheta+1=0`

`sintheta=1/2 color(white)(aaa)sintheta=-1`

For `0<=theta<2pi`, use unit circle to find

`theta =pi/6, (5pi)/6color(white)(aaa)theta =(3pi)/2`

Answer 2

`cos^2theta-sin^2theta=sintheta`

`=>cos2theta=cos(pi/2-theta)`

`2theta=2npi±(pi/2-theta)," where "n in ZZ`

When `2theta=2npi+(pi/2-theta)`

`=>3theta=(4n+1)pi/2`

`=>theta=(4n+1)pi/6`

Again when

`2theta=2npi-(pi/2-theta)`

`=>2theta=2npi-pi/2+theta`

`=>theta=(4n-1)pi/2`