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

The general solution for 2sin(2x+(3pi)/2)+1=0 is x=pin-pi/6 ,x=pin+pi/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(2x+(3pi)/2)+1=0

2 sin (2 x + \frac{3 π}{2}) + 1 = 0

x = πn - \frac{π}{6}, x = πn + \frac{π}{6}

Solution steps

2 sin (2 x + \frac{3 π}{2}) + 1 = 0

Move

1

to the right side

2 sin (2 x + \frac{3 π}{2}) = - 1

Divide both sides by

2

sin (2 x + \frac{3 π}{2}) = - \frac{1}{2}

General solutions for

sin (2 x + \frac{3 π}{2}) = - \frac{1}{2}

2 x + \frac{3 π}{2} = \frac{7 π}{6} + 2 πn, 2 x + \frac{3 π}{2} = \frac{11 π}{6} + 2 πn

Solve

2 x + \frac{3 π}{2} = \frac{7 π}{6} + 2 πn : x = πn - \frac{π}{6}

Solve

2 x + \frac{3 π}{2} = \frac{11 π}{6} + 2 πn : x = πn + \frac{π}{6}

x = πn - \frac{π}{6}, x = πn + \frac{π}{6}

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

2 cos^{2} (θ) - 2 sin^{2} (θ) = 2

cot (θ) + 3 csc (θ) = 5

2 cos^{2} (θ) - 2 sin^{2} (θ) = 3

1 - ∣ cos (x) ∣ = 0

What is the general solution for 2sin(2x+(3pi)/2)+1=0 ?
The general solution for 2sin(2x+(3pi)/2)+1=0 is x=pin-pi/6 ,x=pin+pi/6