What is the general solution for 2sin(1/2 x)=1,0<= x<= 12 ?

The general solution for 2sin(1/2 x)=1,0<= x<= 12 is No Solution for x\in\mathbb{R}

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2sin(1/2 x)=1,0<= x<= 12

2 sin (\frac{1}{2} x) = 1, 0^{\circ} \leq x \leq 1 2^{\circ}

No Solution for x \in R

Solution steps

2 sin (\frac{1}{2} x) = 1, 0^{\circ} \leq x \leq 1 2^{\circ}

Divide both sides by

2

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

General solutions for

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

\frac{1}{2} x = 3 0^{\circ} + 36 0^{\circ} n, \frac{1}{2} x = 15 0^{\circ} + 36 0^{\circ} n

Solve

\frac{1}{2} x = 3 0^{\circ} + 36 0^{\circ} n : x = 6 0^{\circ} + 72 0^{\circ} n

Solve

\frac{1}{2} x = 15 0^{\circ} + 36 0^{\circ} n : x = 30 0^{\circ} + 72 0^{\circ} n

x = 6 0^{\circ} + 72 0^{\circ} n, x = 30 0^{\circ} + 72 0^{\circ} n

Solutions for the range

0 \leq x \leq 1 2^{\circ}

No Solution for x \in R

2 sin (\frac{π}{4} + x) \cdot sin (\frac{π}{4} - x) + sin^{2} (x) = 0

(2 sin (x) - 1) (cos (x) + 1) = 0

7 sin^{2} (x) - 6 = - 3 sin^{2} (x)

tan (b) = 4

1 sin (3 0^{\circ}) = 1.5 sin (x)

What is the general solution for 2sin(1/2 x)=1,0<= x<= 12 ?
The general solution for 2sin(1/2 x)=1,0<= x<= 12 is No Solution for x\in\mathbb{R}