What is the general solution for ((1+tan(x)))/(sec(x))=2,x ?

The general solution for ((1+tan(x)))/(sec(x))=2,x is No Solution for x\in\mathbb{R}

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((1+tan(x)))/(sec(x))=2,x

\frac{( 1 + tan ( x ) )}{sec ( x )} = 2, x

No Solution for x \in R

Solution steps

\frac{( 1 + tan ( x ) )}{sec ( x )} = 2, x

Subtract

2

from both sides

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

Simplify

\frac{1 + tan ( x )}{sec ( x )} - 2 : \frac{1 + tan ( x ) - 2 sec ( x )}{sec ( x )}

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

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

1 + tan (x) - 2 sec (x) = 0

Express with sin, cos

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

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

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

Rewrite using trig identities

- 2 + 2 sin (x + \frac{π}{4}) = 0

Move

2

to the right side

2 sin (x + \frac{π}{4}) = 2

Divide both sides by

2

sin (x + \frac{π}{4}) = 2

- 1 \leq sin (x) \leq 1

No Solution for x \in R

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

cos (θ) = - \frac{7}{15}, cos (\frac{θ}{2}), 18 0^{\circ} < θ < 27 0^{\circ}

tan (x) + 2 = 0

2 6^{2} = 1 3^{2} + 1 9^{2} - (13) (19) cos (J)

4 = sec^{2} (x)

What is the general solution for ((1+tan(x)))/(sec(x))=2,x ?
The general solution for ((1+tan(x)))/(sec(x))=2,x is No Solution for x\in\mathbb{R}