Question

Solve: `5sin theta = 6 cos^2theta -6` for `theta in [0, 2pi]`?

Answer 1

`theta = 0, pi, approx 4.127, approx 5.298,2pi` for `theta in [0,2pi]`

Explanation:

`5sin theta = 6 cos^2theta -6`

`5sin theta = 6(cos^2theta -1)`

`5sin theta = -6sin^2theta`

`sin theta(5+6sin theta) =0`

`sin theta = 0 or sin theta = -5/6`

`sin theta =0 ->theta = 0, pi or 2pi in [0, 2pi]`

`sin theta = -5/6 -> theta = arcsin (-5/6) approx 4.127 or approx 5.298 in [0, 2pi]`

Hence, `theta = 0, pi, approx 4.127, approx 5.298,2pi` for `theta in [0,2pi]`

We can see these results from the zeros of the graph of `sin theta(5+6sin theta)` for `theta in [0,2pi]` below:

graph{sinx(5+6sinx) [-1.56, 7.21, -1.892, 2.49]}

Answer 2

`theta=0; pi; 4.126; 5.298; 2pi`

Explanation:

`5sintheta=6cos^2theta-6`

`5sintheta=6(cos^2theta-1)`

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`1=sin^2x+cos^2xquad=>quadsin^2x=1-cos^2x`

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`5sintheta=-6(1-cos^2theta)`

`5sintheta=-6sin^2theta`

`5sintheta+6sin^2theta=0`

`sintheta(5+6sintheta)=0`

`sintheta=0quad=>quadtheta_1=0quad^^quadtheta_2=piquad^^quadtheta_3=2pi`

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`5+6sintheta_(4,5)=0`
`=>sintheta_(4,5)=-5/6`
To find that you have to use calculator but it will tell you only one value. To find second value we need to know in which quadrant we are.

`=>theta_(4)~~-0.985~~-pi/3.1`
but `0<=theta<=2pi`
`=>theta_4~~-0.985+2pi~~color(pink)5.298~~(5pi)/3`

`(3pi)/2<=(5pi)/3<=2piquad=>quad` 4th quadrant
That means that the last angle must be in 3rd quadrant. To calculate that we use well known trick:
For 3rd quadrant: `varphi=pi+varphi_0`

Applying to our case: `theta_0=|-0.985|`
`=>theta_5=pi+0.985=color(orange)4.126`