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I'm having trouble with this trigonometric equation:

 

Solve sin(3x) = sin(x) for 0 <= x <= 2*pi.  I'd appreciate any help.

 May 10, 2020
 #1
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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.

 May 10, 2020

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