What is the general solution for sin(x/7)=(3^{1/2})/2 ,pi<x<3pi ?

The general solution for sin(x/7)=(3^{1/2})/2 ,pi<x<3pi is x=(7pi)/3

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sin(x/7)=(3^{1/2})/2 ,pi<x<3pi

sin (\frac{x}{7}) = \frac{3 ^{\frac{1}{2}}}{2}, π < x < 3 π

x = \frac{7 π}{3}

Solution steps

sin (\frac{x}{7}) = \frac{3 ^{\frac{1}{2}}}{2}, π < x < 3 π

Apply trig inverse properties

\frac{x}{7} = arcsin (\frac{3 ^{\frac{1}{2}}}{2}) + 2 πn, \frac{x}{7} = π - arcsin (\frac{3 ^{\frac{1}{2}}}{2}) + 2 πn

Solve

\frac{x}{7} = arcsin (\frac{3 ^{\frac{1}{2}}}{2}) + 2 πn : x = \frac{7 π}{3} + 14 πn

Solve

\frac{x}{7} = π - arcsin (\frac{3 ^{\frac{1}{2}}}{2}) + 2 πn : x = 7 π - \frac{7 π}{3} + 14 πn

x = \frac{7 π}{3} + 14 πn, x = 7 π - \frac{7 π}{3} + 14 πn

Solutions for the range

π < x < 3 π

x = \frac{7 π}{3}

2 sec^{2} (x) - 13 sec (x) + 20 = 0, 2 π \leq x \leq 4 π

1 = - sin (x)

cos (4 5^{\circ} - x) = \frac{1}{2}

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

20 sin (x) = 20 cos (x)

What is the general solution for sin(x/7)=(3^{1/2})/2 ,pi<x<3pi ?
The general solution for sin(x/7)=(3^{1/2})/2 ,pi<x<3pi is x=(7pi)/3