Question

Help please?!

Answer 1

See below.

Explanation:

Calling `f(theta) = 6 Cos^2theta - 8 Sin theta Cos theta - (A + R Cos(2 theta + d))` we have

`f(0) = 6 - A - R Cos d = 0`
`f(pi/2) = -A + R Cosd = 0`

now deriving `f(theta)`

`f'(0) = -8+2Rsind=0`

Solving now

`{( 6 - A - R Cos d = 0),( -A + R Cosd = 0),(-8+2Rsind=0):}`

we get

`{(A=3),(R=-5),(d = arctan(4/3)-pi + 2kpi):}`

and

`{(A=3),(R=5),(d = arctan(4/3) + 2kpi):}`

for `k=0,pm1,pm2,pm3,cdots`

so, to solve

`6 Cos^2theta - 8 Sin theta Cos theta=5`

we can use instead

`3+5cos(2theta+phi)=5` or

`cos(2theta+phi)=2/5`

Here `phi = arctan(4/3)` and then

`2theta+phi = arccos(2/5)` and finally

`theta = (1/2arccos(2/5)-phi)`