What is the general solution for (sin(2x))/(cos(x))=1 ?

The general solution for (sin(2x))/(cos(x))=1 is x= pi/6+2pin,x=(5pi)/6+2pin

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(sin(2x))/(cos(x))=1

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

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

Solution steps

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

Subtract

1

from both sides

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

Simplify

\frac{sin ( 2 x )}{cos ( x )} - 1 : \frac{sin ( 2 x ) - cos ( x )}{cos ( x )}

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

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

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

Rewrite using trig identities

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

Factor

- cos (x) + 2 cos (x) sin (x) : cos (x) (2 sin (x) - 1)

cos (x) (2 sin (x) - 1) = 0

Solving each part separately

cos (x) = 0 or 2 sin (x) - 1 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

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

tan^{2} (x) + 4 tan (x) + 1 = 0

sin (c) = \frac{3}{4}

2 cosh^{2} (x) - 5 sinh (x) = 5

10 + 5 cos (2 x) = 15 cos (x)

y, x = cos (2 y)

What is the general solution for (sin(2x))/(cos(x))=1 ?
The general solution for (sin(2x))/(cos(x))=1 is x= pi/6+2pin,x=(5pi)/6+2pin