Question

2 sin x + cotx -cos ecx = 0​

Answer 1

Step-by-step explanation:

Given that:-

2 sin x + cotx -cosecx = 0

2 sin x+(cos x/sin x)-(1/sin x)=0

=>(2 sin² x+ cos x-1)/sin x=0

=>2 sin² x+ cos x-1=0×sin x

=>2 sin²x+cos x-1=0

=>2(1-cos²x)+cos x-1=0

=>2-2cos²x+cos x-1=0

=>-2 Cos²x+Cos x+1=0

=>2 Cos²x-cos x-1=0

=>2 Cos²x-2Cosx+Cosx-1=0

=>2Cos x(Cos x-1)+1(Cos x-1)=0

=>(2 Cos x+1)(Cos x-1)=0

=>2 cos x+1=0 or Cosx-1=0

=>2 Cos x=-1 or Cos x=1

=>Cos x=-1/2 or Cosx =1

Cos x=1

=>Cos x=Cos 0

=>X=0°

Cos x=-1/2

Cos x=Cos 120°

X=120°

If x lies between 0° to 90° then

X=0°

Used concept:-

Sin²A+Cos²A=1

Negative Values of Cos lies in 2nd and 3rd quadrants