Question

Solve for x [0, 2pi]
2sinx=3+2cscx

Answer 1

start with 2sin(x)*sin(x)=3sin(x)+2; hint write cosec(x)=1/sin(x)

Answer 2

2sin^2(x)-3sin(x)-2=0;
2sin^2(x)-4sin(x)+sin(x)-2=0;
2sin(x)[sin(x)-2]+1[sin(x)-2]=0;
sin(x)-2= 0; or 2sin(x)+1=0;

sin(x) IS NOT EQUAL TO 2; so sin(x)=-(1/2),

x belongs to 0 to 2pi, and sin is negative in third and fourth quadrant,

so x= pi+(pi/6) or x= 2pi-(pi/6), solve that

Answer 3

does it help?

Answer 4

how do you get 2sin(x)*sin(x)=3sin(x)+2? because when i try to break it down i get sinx+sinx=3+2/sinx

Answer 5

Write cosec(x)= 1/sin(x);
then take a common denominator or simply put, bring sin(x)on left side after taking LCM(least common multiple or whatever term you prefer to use)

Answer 6

ok, i get it now