Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.) 2 sin x − cot x − csc x = 0

Question

Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)

2 sin x − cot x − csc x = 0

Answer 1

2sinx-cotx-csx=0 The first thing you have to do is convert cot(x )in terms of cos(x)/ sin(x) and csc(x) in terms of 1/sin (x). Let do these substitution to see how the trigonometric equations is transformed:

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

Multiplying each term by the common denominator sin (x) we got:

2 sin^2(x) -cos (x) -1 =0. but I notice that sin^2(x) = 1-cos^x , let substitute this identity now

2(1-cos^2(x)) -cos(x) -1 =0 Now I see that every thing is in term of cos(x).

2-2cos^2(x) -cos (x) -1=0. adding 2 and -1 =1

-2cos^2(x)-cos(x) +1 =0. multiplying each term by -1

2cos^2 (x)+cos (x) -1 =0. Factoring

(2cos. - 1 ) ( cos (x). +1) =0. Applying property of zero

2cos(x) -1. =0.

cos (x) - 1/2=0 The angle x that correspond to 1/2 in the unit circle is 60 degrees or pi/3 and 300 degrees or 5pi/3.

For the second factor (cos(x) +1) =0

cos(x) =-1. The angle that correspond to -1 in the unit circle is 180 degrees or pi.

Therefore the answer to this problem is : pi/3, 5pi/3 and pi. Between 0≤x≤2∏