What is the general solution for sqrt(9cos(2θ))=0 ?

The general solution for sqrt(9cos(2θ))=0 is θ= pi/4+pin,θ=(3pi)/4+pin

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sqrt(9cos(2θ))=0

Solution

9 cos (2 θ) = 0

Solution

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

Solution steps

9 cos (2 θ) = 0

Square both sides:

9 cos (2 θ) = 0

9 cos (2 θ) = 0

(9 cos (2 θ))^{2} = 0^{2}

Expand

(9 cos (2 θ))^{2} : 9 cos (2 θ)

(9 cos (2 θ))^{2}

Apply radical rule:

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

= ((9 cos (2 θ))^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (9 cos (2 θ))^{\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

= 9 cos (2 θ)

Expand

0^{2} : 0

0^{2}

Apply rule

0^{a} = 0

= 0

9 cos (2 θ) = 0

9 cos (2 θ) = 0

Solve

9 cos (2 θ) = 0 : cos (2 θ) = 0

9 cos (2 θ) = 0

Divide both sides by

9

9 cos (2 θ) = 0

Divide both sides by

9

\frac{9 cos ( 2 θ )}{9} = \frac{0}{9}

Simplify

cos (2 θ) = 0

cos (2 θ) = 0

cos (2 θ) = 0

Verify Solutions:

cos (2 θ) = 0

True

Check the solutions by plugging them into

9 cos (2 θ) = 0

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

Plug in

cos (2 θ) = 0 :

True

9 \cdot 0 = 0

9 \cdot 0 = 0

9 \cdot 0

Apply rule

0 \cdot a = 0

= 0

Apply rule

0 = 0

= 0

0 = 0

True

The solution is

cos (2 θ) = 0

General solutions for

cos (2 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}

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

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

Solve

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solve

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

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

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

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(9cos(2θ))=0 ?
The general solution for sqrt(9cos(2θ))=0 is θ= pi/4+pin,θ=(3pi)/4+pin