What is the general solution for 3sqrt(2)cos(θ)+1=-2,0<= θ<2pi ?

Q: What is the general solution for 3sqrt(2)cos(θ)+1=-2,0<= θ<2pi ?

The general solution for 3sqrt(2)cos(θ)+1=-2,0<= θ<2pi is θ=(3pi)/4 ,θ=(5pi)/4

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Popular >

3sqrt(2)cos(θ)+1=-2,0<= θ<2pi

Solution

32 cos (θ) + 1 = - 2, 0 \leq θ < 2 π

Solution

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

Degrees

θ = 13 5^{\circ}, θ = 22 5^{\circ}

Solution steps

32 cos (θ) + 1 = - 2, 0 \leq θ < 2 π

Move

1

to the right side

32 cos (θ) + 1 = - 2

Subtract

1

from both sides

32 cos (θ) + 1 - 1 = - 2 - 1

Simplify

32 cos (θ) = - 3

32 cos (θ) = - 3

Divide both sides by

32

32 cos (θ) = - 3

Divide both sides by

32

\frac{3 2 cos ( θ )}{3 2} = \frac{- 3}{3 2}

Simplify

\frac{3 2 cos ( θ )}{3 2} = \frac{- 3}{3 2}

Simplify

\frac{3 2 cos ( θ )}{3 2} : cos (θ)

\frac{3 2 cos ( θ )}{3 2}

Divide the numbers:

\frac{3}{3} = 1

= \frac{2 cos ( θ )}{2}

Cancel the common factor:

2

= cos (θ)

Simplify

\frac{- 3}{3 2} : - \frac{2}{2}

\frac{- 3}{3 2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{3 2}

Divide the numbers:

\frac{3}{3} = 1

= - \frac{1}{2}

Rationalize

- \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}

= - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \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}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

Graph

Popular Examples

cos(u)=-15/17

cos (u) = - \frac{15}{17}

2cos(x/3)=1

2 cos (\frac{x}{3}) = 1

0=3tan(x)

0 = 3 tan (x)

solvefor x,z=arctan((x-y)/(x+y))solve for

x, z = arctan (\frac{x - y}{x + y})

sin(b)= 4/5

sin (b) = \frac{4}{5}

Frequently Asked Questions (FAQ)

What is the general solution for 3sqrt(2)cos(θ)+1=-2,0<= θ<2pi ?
The general solution for 3sqrt(2)cos(θ)+1=-2,0<= θ<2pi is θ=(3pi)/4 ,θ=(5pi)/4