What is the general solution for sin(θ)=(0.3924)/(cos(θ)) ?

The general solution for sin(θ)=(0.3924)/(cos(θ)) is θ=(0.90237…)/2+pin,θ= pi/2-(0.90237…)/2+pin

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sin(θ)=(0.3924)/(cos(θ))

sin (θ) = \frac{0.3924}{cos ( θ )}

θ = \frac{0.90237 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.90237 \dots}{2} + πn

Solution steps

sin (θ) = \frac{0.3924}{cos ( θ )}

Subtract

\frac{0.3924}{cos ( θ )}

from both sides

sin (θ) - \frac{0.3924}{cos ( θ )} = 0

Simplify

sin (θ) - \frac{0.3924}{cos ( θ )} : \frac{sin ( θ ) cos ( θ ) - 0.3924}{cos ( θ )}

\frac{sin ( θ ) cos ( θ ) - 0.3924}{cos ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (θ) cos (θ) - 0.3924 = 0

Rewrite using trig identities

- 0.3924 + \frac{sin ( 2 θ )}{2} = 0

Move

0.3924

to the right side

\frac{sin ( 2 θ )}{2} = 0.3924

Multiply both sides by

2

sin (2 θ) = 0.7848

Apply trig inverse properties

2 θ = arcsin (0.7848) + 2 πn, 2 θ = π - arcsin (0.7848) + 2 πn

Solve

2 θ = arcsin (0.7848) + 2 πn : θ = \frac{arcsin ( 0.7848 )}{2} + πn

Solve

2 θ = π - arcsin (0.7848) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.7848 )}{2} + πn

θ = \frac{arcsin ( 0.7848 )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.7848 )}{2} + πn

Show solutions in decimal form

θ = \frac{0.90237 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.90237 \dots}{2} + πn

x, y = 3 sin (x)

6 - 4 cos^{2} (x) - 5 cos (x) = 0

cos (x) = \frac{32}{34.54}

tan (a) = \frac{51}{140}

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

What is the general solution for sin(θ)=(0.3924)/(cos(θ)) ?
The general solution for sin(θ)=(0.3924)/(cos(θ)) is θ=(0.90237…)/2+pin,θ= pi/2-(0.90237…)/2+pin