What is the general solution for 4cos(x)-1=sqrt(5) ?

The general solution for 4cos(x)-1=sqrt(5) is x=0.62831…+2pin,x=2pi-0.62831…+2pin

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4cos(x)-1=sqrt(5)

4 cos (x) - 1 = 5

x = 0.62831 \dots + 2 πn, x = 2 π - 0.62831 \dots + 2 πn

Degrees

x = 3 6^{\circ} + 36 0^{\circ} n, x = 32 4^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (x) - 1 = 5

Move

1

to the right side

4 cos (x) - 1 = 5

Add

1

to both sides

4 cos (x) - 1 + 1 = 5 + 1

Simplify

4 cos (x) = 5 + 1

4 cos (x) = 5 + 1

Divide both sides by

4

4 cos (x) = 5 + 1

Divide both sides by

4

\frac{4 cos ( x )}{4} = \frac{5}{4} + \frac{1}{4}

Simplify

\frac{4 cos ( x )}{4} = \frac{5}{4} + \frac{1}{4}

Simplify

\frac{4 cos ( x )}{4} : cos (x)

\frac{4 cos ( x )}{4}

Divide the numbers:

\frac{4}{4} = 1

= cos (x)

Simplify

\frac{5}{4} + \frac{1}{4} : \frac{5 + 1}{4}

\frac{5}{4} + \frac{1}{4}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 + 1}{4}

cos (x) = \frac{5 + 1}{4}

cos (x) = \frac{5 + 1}{4}

cos (x) = \frac{5 + 1}{4}

Apply trig inverse properties

cos (x) = \frac{5 + 1}{4}

General solutions for

cos (x) = \frac{5 + 1}{4}

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

x = arccos (\frac{5 + 1}{4}) + 2 πn, x = 2 π - arccos (\frac{5 + 1}{4}) + 2 πn

x = arccos (\frac{5 + 1}{4}) + 2 πn, x = 2 π - arccos (\frac{5 + 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.62831 \dots + 2 πn, x = 2 π - 0.62831 \dots + 2 πn

cos (4 y) = 2 cos (2 y) - 1

csc (x) cot (x) = 23

sin (θ) cos (θ) - sin (θ) = 0

cos^{2} (θ) + 2 sin (θ) + 1 = 0

arcsec (x) = arcsec (2)

What is the general solution for 4cos(x)-1=sqrt(5) ?
The general solution for 4cos(x)-1=sqrt(5) is x=0.62831…+2pin,x=2pi-0.62831…+2pin