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

The general solution for 4cos(x)=sqrt(5)-1 is x=1.25663…+2pin,x=2pi-1.25663…+2pin

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

4 cos (x) = 5 - 1

x = 1.25663 \dots + 2 πn, x = 2 π - 1.25663 \dots + 2 πn

Degrees

x = 7 2^{\circ} + 36 0^{\circ} n, x = 28 8^{\circ} + 36 0^{\circ} n

Solution steps

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 = 1.25663 \dots + 2 πn, x = 2 π - 1.25663 \dots + 2 πn

2 sin^{2} (θ) - 2 sin (θ) = 0

2 - cos (θ) = 2 + sin (θ)

sin (x) = 6757

\frac{1}{cos ( x )} + tan (x) = 2

- 5 = - 5 + 4 sin (- 2 x + \frac{3 π}{4})

What is the general solution for 4cos(x)=sqrt(5)-1 ?
The general solution for 4cos(x)=sqrt(5)-1 is x=1.25663…+2pin,x=2pi-1.25663…+2pin