What is the general solution for sqrt(2)-4cos(A)=0 ?

The general solution for sqrt(2)-4cos(A)=0 is A=1.20942…+2pin,A=2pi-1.20942…+2pin

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sqrt(2)-4cos(A)=0

2 - 4 cos (A) = 0

A = 1.20942 \dots + 2 πn, A = 2 π - 1.20942 \dots + 2 πn

Degrees

A = 69.29518 \dots^{\circ} + 36 0^{\circ} n, A = 290.70481 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 - 4 cos (A) = 0

Move

2

to the right side

2 - 4 cos (A) = 0

Subtract

2

from both sides

2 - 4 cos (A) - 2 = 0 - 2

Simplify

- 4 cos (A) = - 2

- 4 cos (A) = - 2

Divide both sides by

- 4

- 4 cos (A) = - 2

Divide both sides by

- 4

\frac{- 4 cos ( A )}{- 4} = \frac{- 2}{- 4}

Simplify

cos (A) = \frac{2}{4}

cos (A) = \frac{2}{4}

Apply trig inverse properties

cos (A) = \frac{2}{4}

General solutions for

cos (A) = \frac{2}{4}

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

A = arccos (\frac{2}{4}) + 2 πn, A = 2 π - arccos (\frac{2}{4}) + 2 πn

A = arccos (\frac{2}{4}) + 2 πn, A = 2 π - arccos (\frac{2}{4}) + 2 πn

Show solutions in decimal form

A = 1.20942 \dots + 2 πn, A = 2 π - 1.20942 \dots + 2 πn

6 cos (6 x) = 3

x, sin (x) cos (y) = 0

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

\frac{sin ( 9 5 ^{\circ} )}{27} = \frac{sin ( b )}{16}

cos (x) + \frac{3}{2} = 0, 0 \leq x \leq 2 π

What is the general solution for sqrt(2)-4cos(A)=0 ?
The general solution for sqrt(2)-4cos(A)=0 is A=1.20942…+2pin,A=2pi-1.20942…+2pin