What is the general solution for sin(2x)=(-1)/(sqrt(3)) ?

The general solution for sin(2x)=(-1)/(sqrt(3)) is x=-(0.61547…)/2+pin,x= pi/2+(0.61547…)/2+pin

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sin(2x)=(-1)/(sqrt(3))

sin (2 x) = \frac{- 1}{3}

x = - \frac{0.61547 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.61547 \dots}{2} + πn

Solution steps

sin (2 x) = \frac{- 1}{3}

Simplify

\frac{- 1}{3} : - \frac{3}{3}

Apply trig inverse properties

2 x = arcsin (- \frac{3}{3}) + 2 πn, 2 x = π + arcsin (\frac{3}{3}) + 2 πn

Solve

2 x = arcsin (- \frac{3}{3}) + 2 πn : x = - \frac{arcsin ( \frac{3}{3} )}{2} + πn

Solve

2 x = π + arcsin (\frac{3}{3}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{3}{3} )}{2} + πn

x = - \frac{arcsin ( \frac{3}{3} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{3}{3} )}{2} + πn

Show solutions in decimal form

x = - \frac{0.61547 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.61547 \dots}{2} + πn

cos (2 θ) = 1 + 3 sin (θ)

cos (2 x) - cos (x) = - 1, 0 \leq x < 2 π

sin (x) = - \frac{1}{10}

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

6 sec^{2} (x) = 0

What is the general solution for sin(2x)=(-1)/(sqrt(3)) ?
The general solution for sin(2x)=(-1)/(sqrt(3)) is x=-(0.61547…)/2+pin,x= pi/2+(0.61547…)/2+pin