What is the general solution for sin(2x)=-0.8 ?

The general solution for sin(2x)=-0.8 is x=-(0.92729…)/2+pin,x= pi/2+(0.92729…)/2+pin

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sin(2x)=-0.8

sin (2 x) = - 0.8

x = - \frac{0.92729 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.92729 \dots}{2} + πn

Solution steps

sin (2 x) = - 0.8

Apply trig inverse properties

2 x = arcsin (- 0.8) + 2 πn, 2 x = π + arcsin (0.8) + 2 πn

Solve

2 x = arcsin (- 0.8) + 2 πn : x = - \frac{arcsin ( \frac{4}{5} )}{2} + πn

Solve

2 x = π + arcsin (0.8) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( 0.8 )}{2} + πn

x = - \frac{arcsin ( \frac{4}{5} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( 0.8 )}{2} + πn

Show solutions in decimal form

x = - \frac{0.92729 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.92729 \dots}{2} + πn

cos (2 x - \frac{π}{5}) = 0

cos (a) = sin (a + 3 0^{\circ})

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

\frac{sin ( 5 0 ^{\circ} )}{10} = \frac{sin ( q )}{12}

271.63 = \frac{3 3 \cdot 169.7}{π} cos (α)

What is the general solution for sin(2x)=-0.8 ?
The general solution for sin(2x)=-0.8 is x=-(0.92729…)/2+pin,x= pi/2+(0.92729…)/2+pin