What is the general solution for sin(2x)=-0.848055484 ?

The general solution for sin(2x)=-0.848055484 is x=-(1.01230…)/2+pin,x= pi/2+(1.01230…)/2+pin

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sin(2x)=-0.848055484

sin (2 x) = - 0.848055484

x = - \frac{1.01230 \dots}{2} + πn, x = \frac{π}{2} + \frac{1.01230 \dots}{2} + πn

Solution steps

sin (2 x) = - 0.848055484

Apply trig inverse properties

2 x = arcsin (- 0.848055484) + 2 πn, 2 x = π + arcsin (0.848055484) + 2 πn

Solve

2 x = arcsin (- 0.848055484) + 2 πn : x = - \frac{arcsin ( 0.848055484 )}{2} + πn

Solve

2 x = π + arcsin (0.848055484) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( 0.848055484 )}{2} + πn

x = - \frac{arcsin ( 0.848055484 )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( 0.848055484 )}{2} + πn

Show solutions in decimal form

x = - \frac{1.01230 \dots}{2} + πn, x = \frac{π}{2} + \frac{1.01230 \dots}{2} + πn

2 + cos^{2} (x) = 3 cos (x)

cos (x + 4 0^{\circ}) = 0.85

sin^{2} (a) = 1 - cos (2 a)

cos^{4} (x) = cos^{2} (x)

tan (x) = 2 + tan^{3} (x)

What is the general solution for sin(2x)=-0.848055484 ?
The general solution for sin(2x)=-0.848055484 is x=-(1.01230…)/2+pin,x= pi/2+(1.01230…)/2+pin