What is the general solution for 3sin(2x-1)=1 ?

The general solution for 3sin(2x-1)=1 is x=pin+1/2+(0.33983…)/2 ,x=pin+pi/2+1/2-(0.33983…)/2

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3sin(2x-1)=1

3 sin (2 x - 1) = 1

x = πn + \frac{1}{2} + \frac{0.33983 \dots}{2}, x = πn + \frac{π}{2} + \frac{1}{2} - \frac{0.33983 \dots}{2}

Solution steps

3 sin (2 x - 1) = 1

Divide both sides by

3

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

Apply trig inverse properties

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

Solve

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

Solve

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

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

Show solutions in decimal form

x = πn + \frac{1}{2} + \frac{0.33983 \dots}{2}, x = πn + \frac{π}{2} + \frac{1}{2} - \frac{0.33983 \dots}{2}

m, sec (m + 1 2^{\circ}) = csc (4 2^{\circ} - n)

1 tan (x) = 3

tan (x) = \frac{1.6}{4}

sin^{2} (x) + 7 sin^{2} (x) + 5 cos^{2} (x) + 3 = 0

2 cos (a) = 3 sin^{2} (a)

What is the general solution for 3sin(2x-1)=1 ?
The general solution for 3sin(2x-1)=1 is x=pin+1/2+(0.33983…)/2 ,x=pin+pi/2+1/2-(0.33983…)/2