I'm having trouble with this trigonometric equation:
Solve sin(3x) = sin(x) for 0 <= x <= 2*pi. I'd appreciate any help.
+773 We recall that, sin3x=3sinx−4sin3x , so, the given eqn. becomes,
3sinx−4sin3x+sinx=0 .
4sinx−4sin3x=0 .
4sinx(1−sin2x)=0 .
4sinxcos2x=0 .
sinx=0,cosx=0 .
x=kπ,or,x=(2k+1)π2,k∈Z .
Hence, the Soln. Set is {kπ}∪{(2k+1)π2},k∈Z .
But, x∈[0,2π),&,sinx=0⇒x=0,π , while,
for cosx=0,such x areπ2,3π2 .
^m^ whymenotsmart.
