Question

Find all solutions in the interval [0, 2pi):
Sin 3x = sin x

Answer 1

sin(3x)=sin(2x+x)=sin(x)*cos(2x)+sin(2x)*cos(x)=sin(x)*cos(2x)+2sin(x)cos(x)*cos(x)

sin(x)=0 is a solution so x=k*pi (I) wich k is an integer

sor sinx not equal zero we have

cos(2x)+2cos²(x)=1

since cos²(x)=(1+cos(2x))/2

we have :

cos(2x)+1+cos(2x)=1->cos(2x)=0 -> x=k*pi/2+pi/4 (II)

Answer 2

\[\sin^{-1} \sin(3x) = \sin^{-1} \sin(x)\]
\[3x=x\]
\[2x=0\]
x=0\[x= \pi+k \pi \]
@ganeshie8 could u check my answer i think i didnt made any mistake right ?