What is the general solution for (tan(θ))/(cos(θ))=1 ?

The general solution for (tan(θ))/(cos(θ))=1 is θ=0.66623…+2pin,θ=pi-0.66623…+2pin

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(tan(θ))/(cos(θ))=1

\frac{tan ( θ )}{cos ( θ )} = 1

θ = 0.66623 \dots + 2 πn, θ = π - 0.66623 \dots + 2 πn

Solution steps

\frac{tan ( θ )}{cos ( θ )} = 1

Subtract

1

from both sides

\frac{tan ( θ )}{cos ( θ )} - 1 = 0

Simplify

\frac{tan ( θ )}{cos ( θ )} - 1 : \frac{tan ( θ ) - cos ( θ )}{cos ( θ )}

\frac{tan ( θ ) - cos ( θ )}{cos ( θ )} = 0

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

tan (θ) - cos (θ) = 0

Express with sin, cos

\frac{- cos ^{2} ( θ ) + sin ( θ )}{cos ( θ )} = 0

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

- cos^{2} (θ) + sin (θ) = 0

Rewrite using trig identities

- 1 + sin (θ) + sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{- 1 + 5}{2}, sin (θ) = \frac{- 1 - 5}{2}

sin (θ) = \frac{- 1 + 5}{2} : θ = arcsin (\frac{- 1 + 5}{2}) + 2 πn, θ = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (θ) = \frac{- 1 - 5}{2} :

No Solution

Combine all the solutions

θ = arcsin (\frac{- 1 + 5}{2}) + 2 πn, θ = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

θ = 0.66623 \dots + 2 πn, θ = π - 0.66623 \dots + 2 πn

sin (θ) = - \frac{24}{25}, π < θ < \frac{3 π}{2}

sin (\frac{2 θ}{3}) = 0

3 sec^{2} (x) - 6 = 0

sin (x) = \frac{6}{19}

- 2 sin (x) cos (x) + 3 cos (x) = sin (x) + 3 cos (x)

What is the general solution for (tan(θ))/(cos(θ))=1 ?
The general solution for (tan(θ))/(cos(θ))=1 is θ=0.66623…+2pin,θ=pi-0.66623…+2pin