What is the general solution for tan(2x)+sin(2x)+cos(2x)=sec(2x) ?

The general solution for tan(2x)+sin(2x)+cos(2x)=sec(2x) is x=(pi+2pin)/2 ,x=pin,x=(4pin+pi)/4

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tan(2x)+sin(2x)+cos(2x)=sec(2x)

tan (2 x) + sin (2 x) + cos (2 x) = sec (2 x)

x = \frac{π + 2 πn}{2}, x = πn, x = \frac{4 πn + π}{4}

Solution steps

tan (2 x) + sin (2 x) + cos (2 x) = sec (2 x)

Subtract

sec (2 x)

from both sides

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

Let:

u = 2 x

tan (u) + sin (u) + cos (u) - sec (u) = 0

Express with sin, cos

\frac{- 1 + cos ^{2} ( u ) + sin ( u ) + cos ( u ) sin ( u )}{cos ( u )} = 0

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

- 1 + cos^{2} (u) + sin (u) + cos (u) sin (u) = 0

Rewrite using trig identities

sin (u) - (1 + cos (u)) (1 - cos (u)) + cos (u) sin (u) = 0

Factor

sin (u) - (1 + cos (u)) (1 - cos (u)) + cos (u) sin (u) : (1 + cos (u)) (sin (u) + cos (u) - 1)

(1 + cos (u)) (sin (u) + cos (u) - 1) = 0

Solving each part separately

1 + cos (u) = 0 or sin (u) + cos (u) - 1 = 0

1 + cos (u) = 0 : u = π + 2 πn

sin (u) + cos (u) - 1 = 0 : u = 2 πn, u = 2 πn + \frac{π}{2}

Combine all the solutions

u = π + 2 πn, u = 2 πn, u = 2 πn + \frac{π}{2}

Substitute back

u = 2 x

2 x = π + 2 πn : x = \frac{π + 2 πn}{2}

2 x = 2 πn : x = πn

2 x = 2 πn + \frac{π}{2} : x = \frac{4 πn + π}{4}

x = \frac{π + 2 πn}{2}, x = πn, x = \frac{4 πn + π}{4}

Since the equation is undefined for:

\frac{4 πn + π}{4}

x = \frac{π + 2 πn}{2}, x = πn, x = \frac{4 πn + π}{4}

tan (x) cos^{2} (x) - tan (x) = 0, (0, 2 π)

cos (A) - 1 = 0

4 sin (x) + 7 = 3

1 - cos (4 x) = sin (2 x)

tan (θ) = \frac{2}{1}

What is the general solution for tan(2x)+sin(2x)+cos(2x)=sec(2x) ?
The general solution for tan(2x)+sin(2x)+cos(2x)=sec(2x) is x=(pi+2pin)/2 ,x=pin,x=(4pin+pi)/4