What is the general solution for tan(x)= 1/(sin(2x)),0<= x<= pi ?

The general solution for tan(x)= 1/(sin(2x)),0<= x<= pi is x= pi/4 ,x=(3pi)/4

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tan(x)= 1/(sin(2x)),0<= x<= pi

tan (x) = \frac{1}{sin ( 2 x )}, 0 \leq x \leq π

x = \frac{π}{4}, x = \frac{3 π}{4}

Solution steps

tan (x) = \frac{1}{sin ( 2 x )}, 0 \leq x \leq π

Subtract

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

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = \frac{1}{2}, 0 \leq x \leq π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq π :

No Solution

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}

cos (x) = 9054

1 + cos (2 x) = 2 sin^{2} (x)

6 sin^{2} (\frac{θ}{2}) = cos^{2} (θ) + 5

cos (x) = \frac{1}{4} - sin^{2} (x)

2 sin^{2} (x) - 3 sin (x) cos (x) + cos^{2} (x) = 0

What is the general solution for tan(x)= 1/(sin(2x)),0<= x<= pi ?
The general solution for tan(x)= 1/(sin(2x)),0<= x<= pi is x= pi/4 ,x=(3pi)/4