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

The general solution for (sin(2x))/(sin(x))=1 is x= pi/3+2pin,x=(5pi)/3+2pin

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

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

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

Solution steps

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

Subtract

1

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

2 cos (x) - 1 = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

2 πn, π + 2 πn

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

θ, ma = t sin (θ) - m g

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

\frac{cot ( a )}{cos ( a )} = cos (a)

cos (x) - 2 = 0

tan^{2} (x) + 3 = 3 sec (x)

What is the general solution for (sin(2x))/(sin(x))=1 ?
The general solution for (sin(2x))/(sin(x))=1 is x= pi/3+2pin,x=(5pi)/3+2pin