What is the general solution for sin(2θ)=0.4905 ?

The general solution for sin(2θ)=0.4905 is θ=(0.51266…)/2+pin,θ= pi/2-(0.51266…)/2+pin

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sin(2θ)=0.4905

sin (2 θ) = 0.4905

θ = \frac{0.51266 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.51266 \dots}{2} + πn

Degrees

θ = 14.68672 \dots^{\circ} + 18 0^{\circ} n, θ = 75.31327 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 θ) = 0.4905

Apply trig inverse properties

sin (2 θ) = 0.4905

General solutions for

sin (2 θ) = 0.4905

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

2 θ = arcsin (0.4905) + 2 πn, 2 θ = π - arcsin (0.4905) + 2 πn

2 θ = arcsin (0.4905) + 2 πn, 2 θ = π - arcsin (0.4905) + 2 πn

Solve

2 θ = arcsin (0.4905) + 2 πn : θ = \frac{arcsin ( 0.4905 )}{2} + πn

2 θ = arcsin (0.4905) + 2 πn

Divide both sides by

2

2 θ = arcsin (0.4905) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arcsin ( 0.4905 )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arcsin ( 0.4905 )}{2} + πn

θ = \frac{arcsin ( 0.4905 )}{2} + πn

Solve

2 θ = π - arcsin (0.4905) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.4905 )}{2} + πn

2 θ = π - arcsin (0.4905) + 2 πn

Divide both sides by

2

2 θ = π - arcsin (0.4905) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( 0.4905 )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} - \frac{arcsin ( 0.4905 )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( 0.4905 )}{2} + πn

θ = \frac{arcsin ( 0.4905 )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.4905 )}{2} + πn

Show solutions in decimal form

θ = \frac{0.51266 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.51266 \dots}{2} + πn

tan (θ) = \frac{- 4.86}{14.56}

12 cos (x) + 5 sin (x) = 3

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

2 cos (4 θ) = - 1, 0 \leq θ \leq 2 π

1 \cdot sin (4 5^{\circ}) = 1.57 \cdot sin (θ)

What is the general solution for sin(2θ)=0.4905 ?
The general solution for sin(2θ)=0.4905 is θ=(0.51266…)/2+pin,θ= pi/2-(0.51266…)/2+pin