What is the general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi ?

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

The general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi is a= pi/8 ,a=(7pi)/8 ,a=(9pi)/8 ,a=(15pi)/8

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Popular Trigonometry >

2cos(2a)=sqrt(2),0<= θ<= 2pi

Solution

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

Solution

a = \frac{π}{8}, a = \frac{7 π}{8}, a = \frac{9 π}{8}, a = \frac{15 π}{8}

Degrees

a = 22. 5^{\circ}, a = 157. 5^{\circ}, a = 202. 5^{\circ}, a = 337. 5^{\circ}

Solution steps

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

Divide both sides by

2

2 cos (2 a) = 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (2 a) = \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}

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

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

Solve

2 a = \frac{π}{4} + 2 πn : a = \frac{π}{8} + πn

2 a = \frac{π}{4} + 2 πn

Divide both sides by

2

2 a = \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 a}{2} = \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 a}{2} = \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

a = \frac{π}{8} + πn

a = \frac{π}{8} + πn

a = \frac{π}{8} + πn

Solve

2 a = \frac{7 π}{4} + 2 πn : a = \frac{7 π}{8} + πn

2 a = \frac{7 π}{4} + 2 πn

Divide both sides by

2

2 a = \frac{7 π}{4} + 2 πn

Divide both sides by

2

\frac{2 a}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 a}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2} : \frac{7 π}{8} + πn

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{4}}{2} = \frac{7 π}{8}

\frac{\frac{7 π}{4}}{2}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{7 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{8} + πn

a = \frac{7 π}{8} + πn

a = \frac{7 π}{8} + πn

a = \frac{7 π}{8} + πn

a = \frac{π}{8} + πn, a = \frac{7 π}{8} + πn

Solutions for the range

0 \leq θ \leq 2 π

a = \frac{π}{8}, a = \frac{7 π}{8}, a = \frac{9 π}{8}, a = \frac{15 π}{8}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi ?
The general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi is a= pi/8 ,a=(7pi)/8 ,a=(9pi)/8 ,a=(15pi)/8