What is the general solution for tan(x)+sin(x)=((sin(x)+1))/(cos(x)) ?

The general solution for tan(x)+sin(x)=((sin(x)+1))/(cos(x)) is No Solution for x\in\mathbb{R}

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tan(x)+sin(x)=((sin(x)+1))/(cos(x))

tan (x) + sin (x) = \frac{( sin ( x ) + 1 )}{cos ( x )}

No Solution for x \in R

Solution steps

tan (x) + sin (x) = \frac{( sin ( x ) + 1 )}{cos ( x )}

Subtract

\frac{sin ( x ) + 1}{cos ( x )}

from both sides

tan (x) + sin (x) - \frac{sin ( x ) + 1}{cos ( x )} = 0

Simplify

tan (x) + sin (x) - \frac{sin ( x ) + 1}{cos ( x )} : \frac{tan ( x ) cos ( x ) + sin ( x ) cos ( x ) - sin ( x ) - 1}{cos ( x )}

\frac{tan ( x ) cos ( x ) + sin ( x ) cos ( x ) - sin ( x ) - 1}{cos ( x )} = 0

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

tan (x) cos (x) + sin (x) cos (x) - sin (x) - 1 = 0

Rewrite using trig identities

- 1 + \frac{sin ( 2 x )}{2} = 0

Move

1

to the right side

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

Multiply both sides by

2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution for x \in R

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

- 2 sin (x) + cos (x) = 0

3 cos (\frac{2 π t}{3}) + 10 = 11

cos (θ) = \frac{6}{6}, sin (θ)

9 sec^{2} (x) - 9 = 0

What is the general solution for tan(x)+sin(x)=((sin(x)+1))/(cos(x)) ?
The general solution for tan(x)+sin(x)=((sin(x)+1))/(cos(x)) is No Solution for x\in\mathbb{R}