English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos(θ)=sqrt(2cos(θ))

Solution

3 cos (θ) = 2 cos (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 1.34670 \dots + 2 πn, θ = 2 π - 1.34670 \dots + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 77.16041 \dots^{\circ} + 36 0^{\circ} n, θ = 282.83958 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (θ) = 2 cos (θ)

Solve by substitution

3 cos (θ) = 2 cos (θ)

Let:

cos (θ) = u

3 u = 2 u

3 u = 2 u : u = 0, u = \frac{2}{9}

3 u = 2 u

Square both sides:

9 u^{2} = 2 u

3 u = 2 u

(3 u)^{2} = (2 u)^{2}

Expand

(3 u)^{2} : 9 u^{2}

(3 u)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} u^{2}

3^{2} = 9

= 9 u^{2}

Expand

(2 u)^{2} : 2 u

(2 u)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((2 u)^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (2 u)^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2 u

9 u^{2} = 2 u

9 u^{2} = 2 u

Solve

9 u^{2} = 2 u : u = 0, u = \frac{2}{9}

9 u^{2} = 2 u

Move

2 u

to the left side

9 u^{2} = 2 u

Subtract

2 u

from both sides

9 u^{2} - 2 u = 2 u - 2 u

Simplify

9 u^{2} - 2 u = 0

9 u^{2} - 2 u = 0

Factor

9 u^{2} - 2 u : u (9 u - 2)

9 u^{2} - 2 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 9 uu - 2 u

Factor out common term

u

= u (9 u - 2)

u (9 u - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 9 u - 2 = 0

Solve

9 u - 2 = 0 : u = \frac{2}{9}

9 u - 2 = 0

Move

2

to the right side

9 u - 2 = 0

Add

2

to both sides

9 u - 2 + 2 = 0 + 2

Simplify

9 u = 2

9 u = 2

Divide both sides by

9

9 u = 2

Divide both sides by

9

\frac{9 u}{9} = \frac{2}{9}

Simplify

u = \frac{2}{9}

u = \frac{2}{9}

The solutions to the quadratic equation are:

u = 0, u = \frac{2}{9}

u = 0, u = \frac{2}{9}

Verify Solutions:

u = 0

True

, u = \frac{2}{9}

True

Check the solutions by plugging them into

3 u = 2 u

Remove the ones that don't agree with the equation.

Plug in

u = 0 :

True

3 \cdot 0 = 2 \cdot 0

3 \cdot 0 = 0

3 \cdot 0

Apply rule

0 \cdot a = 0

= 0

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

Apply rule

0 = 0

= 0

0 = 0

True

Plug in

u = \frac{2}{9} :

True

3 (\frac{2}{9}) = 2 (\frac{2}{9})

3 (\frac{2}{9}) = \frac{2}{3}

3 (\frac{2}{9})

Remove parentheses:

(a) = a

= 3 \cdot \frac{2}{9}

Multiply fractions:

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

= \frac{2 \cdot 3}{9}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{9}

Cancel the common factor:

3

= \frac{2}{3}

2 (\frac{2}{9}) = \frac{2}{3}

2 (\frac{2}{9})

Remove parentheses:

(a) = a

= 2 \cdot \frac{2}{9}

Multiply

2 \cdot \frac{2}{9} : \frac{4}{9}

2 \cdot \frac{2}{9}

Multiply fractions:

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

= \frac{2 \cdot 2}{9}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{9}

= \frac{4}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{3}

\frac{2}{3} = \frac{2}{3}

True

The solutions are

u = 0, u = \frac{2}{9}

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = \frac{2}{9}

cos (θ) = 0, cos (θ) = \frac{2}{9}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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

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

cos (θ) = \frac{2}{9} : θ = arccos (\frac{2}{9}) + 2 πn, θ = 2 π - arccos (\frac{2}{9}) + 2 πn

cos (θ) = \frac{2}{9}

Apply trig inverse properties

cos (θ) = \frac{2}{9}

General solutions for

cos (θ) = \frac{2}{9}

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

θ = arccos (\frac{2}{9}) + 2 πn, θ = 2 π - arccos (\frac{2}{9}) + 2 πn

θ = arccos (\frac{2}{9}) + 2 πn, θ = 2 π - arccos (\frac{2}{9}) + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = arccos (\frac{2}{9}) + 2 πn, θ = 2 π - arccos (\frac{2}{9}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 1.34670 \dots + 2 πn, θ = 2 π - 1.34670 \dots + 2 πn

Popular Examples

(26.4575)/(sin(60))=(20)/(sin(B))

\frac{26.4575}{sin ( 6 0 ^{\circ} )} = \frac{20}{sin ( B )}

sin(5/2 θ)=-0.966

sin (\frac{5}{2} θ) = - 0.966

sin^2(x)-2/3 =0

sin^{2} (x) - \frac{2}{3} = 0

0=-8sin(x)

0 = - 8 sin (x)

cos^2(x)+2cos(x)+1=0,(0,2pi)

cos^{2} (x) + 2 cos (x) + 1 = 0, (0, 2 π)