solve 2sin(x-45)=1

**mastermin346** · October 6th 2010, 01:38 AM

Solve the equation $2sin(x-45^0) = 1$ for $0^0<x<360^0$

**Educated** · October 6th 2010, 01:46 AM

Just work your way backwards:

$2\sin(x-45^{\circ}) = 1$

$\sin(x-45^{\circ}) = \frac{1}{2}$

$x-45^{\circ} = \sin^{-1}\left(\frac{1}{2}\right)$

Can you carry on?

**Prove It** · October 6th 2010, 01:48 AM

Solve it the way you normally would

$2\sin{(x - 45^{\circ})} = 1$

$\sin{(x - 45^{\circ})} = \frac{1}{2}$

$x - 45^{\circ} = \left\{30^{\circ}, 180^{\circ} - 30^{\circ}\right\}$

$x - 45^{\circ} = \left\{30^{\circ}, 150^{\circ}\right\}$

$x = \left\{30^{\circ} + 45^{\circ}, 150^{\circ} + 45^{\circ}\right\}$

$x = \left\{75^{\circ}, 195^{\circ}\right\}$ .

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