Question

How do I solve this questions?

Solve the given equation:
1) `tantheta-3cottheta=0`
2) `costheta-sintheta=1`

Answer 1

For the equation `cos(theta)-sin(theta)=1`, the solution is `theta=2kpi` and `-pi/2+2kpi` for integers `k`

Explanation:

The second equation is `cos(theta)-sin(theta)=1`.

Consider the equation `sin(pi/4)cos(theta)-cos(pi/4)sin(theta)=sqrt(2)/2`. Notice that this is equivalent to the previous equation as `sin(pi/4)=cos(pi/4)=sqrt(2)/2`.

Then, using the fact that `sin(alphapmbeta)=sin(alpha)cos(beta)pmcos(alpha)sin(beta)`, we have the equation:
`sin(pi/4-theta)=sqrt(2)/2`.

Now, recall that `sin(x)=sqrt(2)/2` when `x=pi/4+2kpi` and `x=(3pi)/4+2kpi` for integers `k`.

Thus,
`pi/4-theta=pi/4+2kpi`
or
`pi/4-theta=(3pi)/4+2kpi`

Finally, we have `theta=2kpi` and `-pi/2+2kpi` for integers `k`.

Answer 2

For the equation `tan(theta)-3cot(theta)=0`, the solution is `theta=pi/3+kpi` or `theta=(2pi)/3+kpi` for integers `k`.

Explanation:

Consider the first equation `tan(theta)-3cot(theta)=0`. We know that `tan(theta)=1/cot(theta)=sin(theta)/cos(theta)`.

Thus, `sin(theta)/cos(theta)-(3cos(theta))/sin(theta)=0`.

Then, `(sin^2(theta)-3cos^2(theta))/(sin(theta)cos(theta))=0`.

Now, if `sin(theta)cos(theta)≠0`, we can safely multiply both sides by `sin(theta)cos(theta)`. This leaves the equation:
`sin^2(theta)-3color(red)(cos^2(theta))=0`

Now, use the identity `cos^2(theta)=color(red)(1-sin^2(theta))` into the red part of the equation above. Substituting this in gives us:
`sin^2(theta)-3(color(red)(1-sin^2(theta)))=0`
`4sin^2(theta)-3=0`
`sin^2(theta)=3/4`
`sin(theta)=pmsqrt(3)/2`

The solution is thus `theta=pi/3+kpi` or `theta=(2pi)/3+kpi` for integers `k`.

(Recall that we required `sin(theta)cos(theta)≠0`. None of the solutions above would give us `sin(theta)cos(theta)=0`, so we're fine here.)