What is the general solution for 4sin(x)+8=8 ?

The general solution for 4sin(x)+8=8 is x=2pin,x=pi+2pin

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4sin(x)+8=8

4 sin (x) + 8 = 8

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (x) + 8 = 8

Subtract

8

from both sides

4 sin (x) + 8 - 8 = 8 - 8

Simplify

4 sin (x) = 0

Divide both sides by

4

4 sin (x) = 0

Divide both sides by

4

\frac{4 sin ( x )}{4} = \frac{0}{4}

Simplify

sin (x) = 0

sin (x) = 0

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

2 cos^{2} (t) - 8 sin^{2} (t) cos (t) - 1 = 0

sin (B) = \frac{4}{5}

cos (2 x) = - 3 cos (x) - 2, 0 < x \leq 2 π

cos (θ) = - \frac{7}{6}

csc (θ) = 5, sin (θ) = x

What is the general solution for 4sin(x)+8=8 ?
The general solution for 4sin(x)+8=8 is x=2pin,x=pi+2pin