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

The general solution for sqrt(2)sin(x)=sqrt(1+(cos(2x))/(sin(x))) is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Subtract

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

from both sides

2 sin (x) = 0

Divide both sides by

2

2 sin (x) = 0

Divide both sides by

2

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

Simplify

sin (x) = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for x \in R

sin (x) = \frac{21}{5}

tan (\frac{θ}{2} - \frac{π}{6}) = 1

1 + 2 sin (θ) = csc (θ)

- 6 = - 4 + sec (x + \frac{π}{4})

2 sin^{2} (x) + 3 sin (x) = 2

What is the general solution for sqrt(2)sin(x)=sqrt(1+(cos(2x))/(sin(x))) ?
The general solution for sqrt(2)sin(x)=sqrt(1+(cos(2x))/(sin(x))) is No Solution for x\in\mathbb{R}