What is the general solution for 4sin(y)= 1/(cos(y)) ?

The general solution for 4sin(y)= 1/(cos(y)) is y= pi/(12)+pin,y=(5pi)/(12)+pin

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4sin(y)= 1/(cos(y))

4 sin (y) = \frac{1}{cos ( y )}

y = \frac{π}{12} + πn, y = \frac{5 π}{12} + πn

Solution steps

4 sin (y) = \frac{1}{cos ( y )}

Subtract

\frac{1}{cos ( y )}

from both sides

4 sin (y) - \frac{1}{cos ( y )} = 0

Simplify

4 sin (y) - \frac{1}{cos ( y )} : \frac{4 sin ( y ) cos ( y ) - 1}{cos ( y )}

\frac{4 sin ( y ) cos ( y ) - 1}{cos ( y )} = 0

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

4 sin (y) cos (y) - 1 = 0

Rewrite using trig identities

- 1 + 4 \cdot \frac{sin ( 2 y )}{2} = 0

4 \cdot \frac{sin ( 2 y )}{2} = 2 sin (2 y)

- 1 + 2 sin (2 y) = 0

Move

1

to the right side

2 sin (2 y) = 1

Divide both sides by

2

sin (2 y) = \frac{1}{2}

General solutions for

sin (2 y) = \frac{1}{2}

2 y = \frac{π}{6} + 2 πn, 2 y = \frac{5 π}{6} + 2 πn

Solve

2 y = \frac{π}{6} + 2 πn : y = \frac{π}{12} + πn

Solve

2 y = \frac{5 π}{6} + 2 πn : y = \frac{5 π}{12} + πn

y = \frac{π}{12} + πn, y = \frac{5 π}{12} + πn

2 cos (x) sin (x) = sin (x)

2 cos (θ) = 2 sin (θ)

arcsin (x) = \frac{π}{8}

5 tan (x) = 8

3 sin (b) + 3 = 5 sin (b) + 2

What is the general solution for 4sin(y)= 1/(cos(y)) ?
The general solution for 4sin(y)= 1/(cos(y)) is y= pi/(12)+pin,y=(5pi)/(12)+pin