What is the general solution for 812.25=399.6001+400-1140cos(a) ?

The general solution for 812.25=399.6001+400-1140cos(a) is a=1.58189…+2pin,a=-1.58189…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

812.25=399.6001+400-1140cos(a)

812.25 = 399.6001 + 400 - 1140 cos (a)

a = 1.58189 \dots + 2 πn, a = - 1.58189 \dots + 2 πn

Degrees

a = 90.63579 \dots^{\circ} + 36 0^{\circ} n, a = - 90.63579 \dots^{\circ} + 36 0^{\circ} n

Solution steps

812.25 = 399.6001 + 400 - 1140 cos (a)

Switch sides

399.6001 + 400 - 1140 cos (a) = 812.25

Add the numbers:

399.6001 + 400 = 799.6001

- 1140 cos (a) + 799.6001 = 812.25

Move

799.6001

to the right side

- 1140 cos (a) + 799.6001 = 812.25

Subtract

799.6001

from both sides

- 1140 cos (a) + 799.6001 - 799.6001 = 812.25 - 799.6001

Simplify

- 1140 cos (a) = 12.6499

- 1140 cos (a) = 12.6499

Divide both sides by

- 1140

- 1140 cos (a) = 12.6499

Divide both sides by

- 1140

\frac{- 1140 cos ( a )}{- 1140} = \frac{12.6499}{- 1140}

Simplify

cos (a) = - \frac{12.6499}{1140}

cos (a) = - \frac{12.6499}{1140}

Apply trig inverse properties

cos (a) = - \frac{12.6499}{1140}

General solutions for

cos (a) = - \frac{12.6499}{1140}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

a = arccos (- \frac{12.6499}{1140}) + 2 πn, a = - arccos (- \frac{12.6499}{1140}) + 2 πn

a = arccos (- \frac{12.6499}{1140}) + 2 πn, a = - arccos (- \frac{12.6499}{1140}) + 2 πn

Show solutions in decimal form

a = 1.58189 \dots + 2 πn, a = - 1.58189 \dots + 2 πn

- 0.6 = sin (2 x)

x, (sin (π x) + cos (π y))^{5} = 36

sin (θ) = - \frac{2}{2}, 0 \leq θ < 2 π

2 sin (\frac{1}{2} x) + 2 = 0

\frac{cos ( x )}{sin ( x )} = - 0.05467

What is the general solution for 812.25=399.6001+400-1140cos(a) ?
The general solution for 812.25=399.6001+400-1140cos(a) is a=1.58189…+2pin,a=-1.58189…+2pin