What is the general solution for sin(2θ)=-(2sqrt(6))/7 ,pi<= θ<= (3pi)/2 ?

The general solution for sin(2θ)=-(2sqrt(6))/7 ,pi<= θ<= (3pi)/2 is No Solution for θ\in\mathbb{R}

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sin(2θ)=-(2sqrt(6))/7 ,pi<= θ<= (3pi)/2

sin (2 θ) = - \frac{2 6}{7}, π \leq θ \leq \frac{3 π}{2}

No Solution for θ \in R

Solution steps

sin (2 θ) = - \frac{2 6}{7}, π \leq θ \leq \frac{3 π}{2}

Apply trig inverse properties

sin (2 θ) = - \frac{2 6}{7}

General solutions for

sin (2 θ) = - \frac{2 6}{7}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

2 θ = arcsin (- \frac{2 6}{7}) + 2 πn, 2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn

2 θ = arcsin (- \frac{2 6}{7}) + 2 πn, 2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn

Solve

2 θ = arcsin (- \frac{2 6}{7}) + 2 πn : θ = - \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

2 θ = arcsin (- \frac{2 6}{7}) + 2 πn

Simplify

arcsin (- \frac{2 6}{7}) + 2 πn : - arcsin (\frac{2 6}{7}) + 2 πn

arcsin (- \frac{2 6}{7}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2 6}{7}) = - arcsin (\frac{2 6}{7})

= - arcsin (\frac{2 6}{7}) + 2 πn

2 θ = - arcsin (\frac{2 6}{7}) + 2 πn

Divide both sides by

2

2 θ = - arcsin (\frac{2 6}{7}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = - \frac{arcsin ( \frac{2 6}{7} )}{2} + \frac{2 πn}{2}

Simplify

θ = - \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

θ = - \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

Solve

2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn

Divide both sides by

2

2 θ = π + arcsin (\frac{2 6}{7}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

θ = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

θ = - \frac{arcsin ( \frac{2 6}{7} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{2 6}{7} )}{2} + πn

Solutions for the range

π \leq θ \leq \frac{3 π}{2}

No Solution for θ \in R

x, cos (x) = \frac{3}{2}

cos (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

cos (2 t) = \frac{3}{2}

cos (x) = - 0.3178, 0 \leq x \leq 2 π

arccos (x) = \frac{π}{3}

What is the general solution for sin(2θ)=-(2sqrt(6))/7 ,pi<= θ<= (3pi)/2 ?
The general solution for sin(2θ)=-(2sqrt(6))/7 ,pi<= θ<= (3pi)/2 is No Solution for θ\in\mathbb{R}