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

The general solution for sqrt(2)cos(x)-1=0,0<= x<= 2pi is x= pi/4 ,x=(7pi)/4

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sqrt(2)cos(x)-1=0,0<= x<= 2pi

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{7 π}{4}

Degrees

x = 4 5^{\circ}, x = 31 5^{\circ}

Solution steps

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

2 cos (x) - 1 + 1 = 0 + 1

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 cos ( x )}{2} : cos (x)

\frac{2 cos ( x )}{2}

Cancel the common factor:

2

= cos (x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (x) = \frac{2}{2}

cos (x) = \frac{2}{2}

cos (x) = \frac{2}{2}

General solutions for

cos (x) = \frac{2}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{7 π}{4}

18 cos (x) + 9 sin (2 x) = 0

sin (x) = \frac{5}{29}

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

sin (a) = \frac{3}{5}, cos (a)

cos (x) = cos (4 x)

What is the general solution for sqrt(2)cos(x)-1=0,0<= x<= 2pi ?
The general solution for sqrt(2)cos(x)-1=0,0<= x<= 2pi is x= pi/4 ,x=(7pi)/4