What is the general solution for 4cos(x)=1,0<= x<= 2pi ?

The general solution for 4cos(x)=1,0<= x<= 2pi is x=1.31811…,x=2pi-1.31811…

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4cos(x)=1,0<= x<= 2pi

4 cos (x) = 1, 0 \leq x \leq 2 π

x = 1.31811 \dots, x = 2 π - 1.31811 \dots

Degrees

x = 75.52248 \dots^{\circ}, x = 284.47751 \dots^{\circ}

Solution steps

4 cos (x) = 1, 0 \leq x \leq 2 π

Divide both sides by

4

4 cos (x) = 1

Divide both sides by

4

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Solutions for the range

0 \leq x \leq 2 π

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

Show solutions in decimal form

x = 1.31811 \dots, x = 2 π - 1.31811 \dots

2 cos (2 x) + 4 sin (x) + 1 = 0

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

sin (x) - 2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

cos (x + π) - cos (x) - 1 = 0

sin (x) = \frac{15}{17}

What is the general solution for 4cos(x)=1,0<= x<= 2pi ?
The general solution for 4cos(x)=1,0<= x<= 2pi is x=1.31811…,x=2pi-1.31811…