Question 546250
{{{4sin(x)-2csc(x)=0}}}
{{{csc(x)=1/sin(x)}}}
so the equation really is
{{{4sin(x)-2/sin(x)=0}}}
To save some writing, let's change variables for a little while. Let's call it {{{y}}}, {{{y=sinx}}}
The equation can be written as
{{{4y-2/y=0}}}
Multiplying times y
{{{4y^2-2=0}}} <--> {{{4y^2=2}}}
The solutions would be {{{y=sqrt(2)/2}}} and {{{y=-sqrt(2)/2}}}
You must know that between 0 and {{{2pi}}}, there are 4 angles that give those values for sin(x):
{{{sin(pi/4)=sin(3pi/4)=sqrt(2)/2}}}
and {{{sin(5pi/4)=sin(7pi/4)=-sqrt(2)/2}}}
To cover all infinity of solutions {{{x=(2k+1)pi/4}}} with k=any integer.