Question

Solve (2+sqrt3)cos theta=1-sin theta?

`(2+sqrt3)cos theta=1-sin theta`

Answer 1

`rarrx=(6n-1)*(pi/3)`

`rarrx=(4n+1)pi/2` Where `nrarrZ`

Explanation:

`rarr(2+sqrt(3))cosx=1-sinx`

`rarrtan75^@*cosx+sinx=1`

`rarr(sin75^@*cosx)/(cos75^@)+sinx=1`

`rarrsinx*cos75^@+cosx*sin75^@=cos75^@=sin(90^@-15^@)=sin15^@`

`rarrsin(x+75^@)-sin15^@=0`

`rarr2sin((x+75^@-15^@)/2)cos((x+75^@+15^@)/2)=0`

`rarrsin((x+60^@)/2)*cos((x+90^@)/2)=0`

Either `rarrsin((x+60^@)/2)=0`

`rarr(x+60^@)/2=npi`

`rarrx=2npi-60^@=2npi-pi/3=(6n-1)*(pi/3)`

or, `cos((x+90^@)/2)=0`

`rarr(x+90^@)/2=(2n+1)pi/2`

`rarrx=2*(2n+1)pi/2-pi/2=(4n+1)pi/2`

Answer 2

If, `costheta=0=>sintheta=1=>theta=(4k+1)pi/2,kinZ`
`theta=2kpi-pi/3,kinZ`,

Explanation:

`(2+sqrt3)costheta=1-sintheta`
`andcostheta!=0`, dividing both sides by `costheta`
`2+sqrt3=sectheta-tantheta=>sectheta-tantheta=2+sqrt3 to (I)`
`:.1/(sectheta-tantheta)=1/(2+sqrt3)` `=>(sec^2theta-tan^2theta)/(sectheta-tantheta)=1/(2+sqrt3)*(2-sqrt3)/(2-sqrt3)`
`=>sectheta+tantheta=2-sqrt3 to (II)`
Adding `(I) and (II)`,we get.`2sectheta=4=>sectheta=2`
`color(red)(costheta=1/2>0)`, From the given equn.
`costheta=1/2=>(2+sqrt3)(1/2)=1-sintheta` `=>1+sqrt(3)/2=1-sintheta=>color(red)(sintheta=-sqrt(3)/2<0)`
`theta=2kpi-pi/3,kinZ`,.......... `(IV^(th)`quadrant)