What is the general solution for cos(2α)+sin(α)=0 ?

The general solution for cos(2α)+sin(α)=0 is α=(7pi)/6+2pin,α=(11pi)/6+2pin,α= pi/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2α)+sin(α)=0

cos (2 α) + sin (α) = 0

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

Solution steps

cos (2 α) + sin (α) = 0

Rewrite using trig identities

1 + sin (α) - 2 sin^{2} (α) = 0

Solve by substitution

sin (α) = - \frac{1}{2}, sin (α) = 1

sin (α) = - \frac{1}{2} : α = \frac{7 π}{6} + 2 πn, α = \frac{11 π}{6} + 2 πn

sin (α) = 1 : α = \frac{π}{2} + 2 πn

Combine all the solutions

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

sin (9 0^{\circ} - x^{\circ}) = \frac{4}{5}

sin (x) = sin (8 0^{\circ})

tan (A) = 2.0503

cot (x) = - 3, 0 \leq x \leq 2 π

2 sin (4 x) - 1 = 0

What is the general solution for cos(2α)+sin(α)=0 ?
The general solution for cos(2α)+sin(α)=0 is α=(7pi)/6+2pin,α=(11pi)/6+2pin,α= pi/2+2pin