What is the general solution for (8cos(x)-4)(sin(x))=0,0<= x<2pi ?

The general solution for (8cos(x)-4)(sin(x))=0,0<= x<2pi is x= pi/3 ,x=(5pi)/3 ,x=0,x=pi

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(8cos(x)-4)(sin(x))=0,0<= x<2pi

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}, x = 0, x = π

Degrees

x = 6 0^{\circ}, x = 30 0^{\circ}, x = 0^{\circ}, x = 18 0^{\circ}

Solution steps

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

Solving each part separately

8 cos (x) - 4 = 0 or sin (x) = 0

8 cos (x) - 4 = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

8 cos (x) - 4 = 0, 0 \leq x < 2 π

Move

4

to the right side

8 cos (x) - 4 = 0

Add

4

to both sides

8 cos (x) - 4 + 4 = 0 + 4

Simplify

8 cos (x) = 4

8 cos (x) = 4

Divide both sides by

8

8 cos (x) = 4

Divide both sides by

8

\frac{8 cos ( x )}{8} = \frac{4}{8}

Simplify

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

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

General solutions for

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

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

sin (x) = 0, 0 \leq x < 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 < 2 π

x = 0, x = π

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}, x = 0, x = π

sin (x) = 3511

tan^{2} (x) - 1 = 4 cos (x)

\frac{10}{sin ( 3 0 ^{\circ} )} = \frac{3}{sin ( b )}

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

arctan (x + 2) - arctan (x + 1) = \frac{π}{4}

What is the general solution for (8cos(x)-4)(sin(x))=0,0<= x<2pi ?
The general solution for (8cos(x)-4)(sin(x))=0,0<= x<2pi is x= pi/3 ,x=(5pi)/3 ,x=0,x=pi