How do I solve this questions?
Solve the given equation:
1) #tantheta-3cottheta=0#
2) #costheta-sintheta=1#
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For the equation #cos(theta)-sin(theta)=1#, the solution is #theta=2kpi# and #-pi/2+2kpi# for integers #k#
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#.
For the equation #tan(theta)-3cot(theta)=0#, the solution is #theta=pi/3+kpi# or #theta=(2pi)/3+kpi# for integers #k#.
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.)
