What is the general solution for (sqrt(3))/2 =sin(arcsin(0)+c_{1}) ?

The general solution for (sqrt(3))/2 =sin(arcsin(0)+c_{1}) is c_{1}= pi/3+2pin,c_{1}=(2pi)/3+2pin

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(sqrt(3))/2 =sin(arcsin(0)+c_{1})

\frac{3}{2} = sin (arcsin (0) + c_{1})

c_{1} = \frac{π}{3} + 2 πn, c_{1} = \frac{2 π}{3} + 2 πn

Solution steps

\frac{3}{2} = sin (arcsin (0) + c_{1})

Switch sides

sin (arcsin (0) + c_{1}) = \frac{3}{2}

arcsin (0) = 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

arcsin (0)

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

= 0

sin (0 + c_{1}) = \frac{3}{2}

General solutions for

sin (0 + c_{1}) = \frac{3}{2}

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}

0 + c_{1} = \frac{π}{3} + 2 πn, 0 + c_{1} = \frac{2 π}{3} + 2 πn

0 + c_{1} = \frac{π}{3} + 2 πn, 0 + c_{1} = \frac{2 π}{3} + 2 πn

Solve

0 + c_{1} = \frac{π}{3} + 2 πn : c_{1} = \frac{π}{3} + 2 πn

0 + c_{1} = \frac{π}{3} + 2 πn

0 + c_{1} = c_{1}

c_{1} = \frac{π}{3} + 2 πn

Solve

0 + c_{1} = \frac{2 π}{3} + 2 πn : c_{1} = \frac{2 π}{3} + 2 πn

0 + c_{1} = \frac{2 π}{3} + 2 πn

0 + c_{1} = c_{1}

c_{1} = \frac{2 π}{3} + 2 πn

c_{1} = \frac{π}{3} + 2 πn, c_{1} = \frac{2 π}{3} + 2 πn

tan^{2} (x) - sin^{2} (x) = tan^{2} (x) + sin^{2} (x)

3 cos (2 x) + cos (x) = 0

sin (2 v) = sin (1 6^{\circ})

tan (2 x) = \frac{130}{- 270 - 120}

- 3 sin (x) - 2 cos (2 x) = 0

What is the general solution for (sqrt(3))/2 =sin(arcsin(0)+c_{1}) ?
The general solution for (sqrt(3))/2 =sin(arcsin(0)+c_{1}) is c_{1}= pi/3+2pin,c_{1}=(2pi)/3+2pin