What is the general solution for 2sqrt(3)*sin(4x+60^0)-3=0 ?

The general solution for 2sqrt(3)*sin(4x+60^0)-3=0 is x=-1/4+pi/(12)+(pin)/2 ,x=-1/4+pi/6+(pin)/2

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2sqrt(3)*sin(4x+60^0)-3=0

23 \cdot sin (4 x + 6 0^{0}) - 3 = 0

x = - \frac{1}{4} + \frac{π}{12} + \frac{πn}{2}, x = - \frac{1}{4} + \frac{π}{6} + \frac{πn}{2}

Solution steps

23 sin (4 x + 6 0^{0}) - 3 = 0

Move

3

to the right side

23 sin (4 x + 6 0^{0}) = 3

Divide both sides by

23

sin (4 x + 6 0^{0}) = \frac{3}{2}

General solutions for

sin (4 x + 6 0^{0}) = \frac{3}{2}

4 x + 6 0^{0} = \frac{π}{3} + 2 πn, 4 x + 6 0^{0} = \frac{2 π}{3} + 2 πn

Solve

4 x + 6 0^{0} = \frac{π}{3} + 2 πn : x = - \frac{1}{4} + \frac{π}{12} + \frac{πn}{2}

Solve

4 x + 6 0^{0} = \frac{2 π}{3} + 2 πn : x = - \frac{1}{4} + \frac{π}{6} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{π}{12} + \frac{πn}{2}, x = - \frac{1}{4} + \frac{π}{6} + \frac{πn}{2}

\frac{sin ( x ) + sin ^{2} ( x )}{2} = 0.5

cos (b) = \frac{3}{5}

arctan (1 - x) + arctan (1 + x) = arctan (\frac{1}{8})

5 sin (4 x) = 2

2 cos^{2} (x) - 3 \cdot sin^{2} (x) - 2 = 0

What is the general solution for 2sqrt(3)*sin(4x+60^0)-3=0 ?
The general solution for 2sqrt(3)*sin(4x+60^0)-3=0 is x=-1/4+pi/(12)+(pin)/2 ,x=-1/4+pi/6+(pin)/2