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

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

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2sin(x)cos(x)=0,0<= x<= 2pi

2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π, x = \frac{π}{2}, x = \frac{3 π}{2}

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 9 0^{\circ}, x = 27 0^{\circ}

Solution steps

2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

Solving each part separately

sin (x) = 0 or cos (x) = 0

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x \leq 2 π

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{π}{2}, x = \frac{3 π}{2}

tan (θ) = - \frac{3}{4}, sec (θ)

\frac{12}{sin ( 12 0 ^{\circ} )} = \frac{5}{sin ( x )}

- 3 tan^{2} (x) = - 1

3 - tan^{2} (b) = 0

\frac{( 2 sin ( x ) ) ( sin ( x ) )}{sin ( x )} + 1 = 3

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