What is the general solution for sin(2x)csc(x)=1,0<= x<= 2pi ?

The general solution for sin(2x)csc(x)=1,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2x)csc(x)=1,0<= x<= 2pi

sin (2 x) csc (x) = 1, 0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

sin (2 x) csc (x) = 1, 0 \leq x \leq 2 π

Subtract

1

from both sides

sin (2 x) csc (x) - 1 = 0

Rewrite using trig identities

- 1 + 2 cos (x) = 0

Move

1

to the right side

2 cos (x) = 1

Divide both sides by

2

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

tan (b) = 3

sin (x) = \frac{12}{16}

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

tan (a) = \frac{3}{2}

cos (x + \frac{π}{6}) = - 1

What is the general solution for sin(2x)csc(x)=1,0<= x<= 2pi ?
The general solution for sin(2x)csc(x)=1,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3