What is the general solution for (-1)e^{-1}cos(y)=0 ?

The general solution for (-1)e^{-1}cos(y)=0 is y= pi/2+2pin,y=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(-1)e^{-1}cos(y)=0

(- 1) e^{- 1} cos (y) = 0

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

Solution steps

(- 1) e^{- 1} cos (y) = 0

Simplify

(- 1) e^{- 1} : - \frac{1}{e}

(- 1) e^{- 1}

Remove parentheses:

(- a) = - a

= - 1 \cdot e^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

e^{- 1} = \frac{1}{e}

= - 1 \cdot \frac{1}{e}

Multiply:

1 \cdot \frac{1}{e} = \frac{1}{e}

= - \frac{1}{e}

(- \frac{1}{e}) cos (y) = 0

Multiply both sides by

- e

(- \frac{1}{e}) cos (y) = 0

Multiply both sides by

- e

(- \frac{1}{e}) cos (y) (- e) = 0 \cdot (- e)

Simplify

cos (y) = 0

cos (y) = 0

General solutions for

cos (y) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

sin (x + 2^{\circ}) = cos (x - 2^{\circ})

cos (x) = \frac{2}{6}

0 = - 6 (sin (t)) + 7 (cos (t))

sec (θ) - 2 = 0, 0 \leq θ \leq 2 π

1 + cot^{2} (x) = csc (x)

What is the general solution for (-1)e^{-1}cos(y)=0 ?
The general solution for (-1)e^{-1}cos(y)=0 is y= pi/2+2pin,y=(3pi)/2+2pin