What is the general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) ?

The general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) is x=(5pi)/4+2pin,x=(7pi)/4+2pin

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(sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2)

\frac{sin ( 2 x )}{sin ( \frac{7 π}{2} - x )} = 2

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solution steps

\frac{sin ( 2 x )}{sin ( \frac{7 π}{2} - x )} = 2

Rewrite using trig identities

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

Subtract

2

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

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

2 + 2 sin (x) = 0 : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Since the equation is undefined for:

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

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

- 2 sec (θ) + 23 = 0, 0^{\circ} \leq θ \leq 36 0^{\circ}

cos (x + 1) = - \frac{2}{5}

4 cos (2 x) + 1 = 3

sinh^{2} (x) = 0

cos (x) csc^{2} (x) + 3 cos (x) = 7 cos (x)

What is the general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) ?
The general solution for (sin(2x)}{sin(\frac{7pi)/2-x)}=sqrt(2) is x=(5pi)/4+2pin,x=(7pi)/4+2pin