What is the general solution for 2sin(A)-1=0 ?

The general solution for 2sin(A)-1=0 is A= pi/6+2pin,A=(5pi)/6+2pin

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2sin(A)-1=0

2 sin (A) - 1 = 0

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

Degrees

A = 3 0^{\circ} + 36 0^{\circ} n, A = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (A) - 1 = 0

Move

1

to the right side

2 sin (A) - 1 = 0

Add

1

to both sides

2 sin (A) - 1 + 1 = 0 + 1

Simplify

2 sin (A) = 1

2 sin (A) = 1

Divide both sides by

2

2 sin (A) = 1

Divide both sides by

2

\frac{2 sin ( A )}{2} = \frac{1}{2}

Simplify

sin (A) = \frac{1}{2}

sin (A) = \frac{1}{2}

General solutions for

sin (A) = \frac{1}{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}

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

A = \frac{π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn

cos (5 θ) = cos (3 θ)

cos (x 1) - cos (x 2) = cos (x 1 - x 2)

8 \cdot sin^{2} (2 x) - 2 \cdot sin (2 x) - 1 = 0

4 - 2 cos (2 x) = 0

x, 2 cos^{2} (x) + sin (x) = 1

What is the general solution for 2sin(A)-1=0 ?
The general solution for 2sin(A)-1=0 is A= pi/6+2pin,A=(5pi)/6+2pin