What is the general solution for 1+sin(2x)=0 ?

The general solution for 1+sin(2x)=0 is x=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1+sin(2x)=0

1 + sin (2 x) = 0

x = \frac{3 π}{4} + πn

Solution steps

1 + sin (2 x) = 0

Move

1

to the right side

sin (2 x) = - 1

General solutions for

sin (2 x) = - 1

2 x = \frac{3 π}{2} + 2 πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

sin^{2} (x) = 2 - cos (x)

cos (2 x) = \frac{2}{2}, 0 \leq x \leq 2 π

sin (2 x) cos (x) = 0

8 cos (x) + 8 sin (x) tan (x) = 16

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