What is the general solution for 2cos(7x)-sqrt(3)=0 ?

The general solution for 2cos(7x)-sqrt(3)=0 is x= pi/(42)+(2pin)/7 ,x=(11pi}{42}+\frac{2pin)/7

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Popular Trigonometry >

2cos(7x)-sqrt(3)=0

Solution

2 cos (7 x) - 3 = 0

Solution

x = \frac{π}{42} + \frac{2 πn}{7}, x = \frac{11 π}{42} + \frac{2 πn}{7}

Degrees

x = 4.28571 \dots^{\circ} + 51.42857 \dots^{\circ} n, x = 47.14285 \dots^{\circ} + 51.42857 \dots^{\circ} n

Solution steps

2 cos (7 x) - 3 = 0

Move

3

to the right side

2 cos (7 x) - 3 = 0

Add

3

to both sides

2 cos (7 x) - 3 + 3 = 0 + 3

Simplify

2 cos (7 x) = 3

2 cos (7 x) = 3

Divide both sides by

2

2 cos (7 x) = 3

Divide both sides by

2

\frac{2 cos ( 7 x )}{2} = \frac{3}{2}

Simplify

cos (7 x) = \frac{3}{2}

cos (7 x) = \frac{3}{2}

General solutions for

cos (7 x) = \frac{3}{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}

7 x = \frac{π}{6} + 2 πn, 7 x = \frac{11 π}{6} + 2 πn

7 x = \frac{π}{6} + 2 πn, 7 x = \frac{11 π}{6} + 2 πn

Solve

7 x = \frac{π}{6} + 2 πn : x = \frac{π}{42} + \frac{2 πn}{7}

7 x = \frac{π}{6} + 2 πn

Divide both sides by

7

7 x = \frac{π}{6} + 2 πn

Divide both sides by

7

\frac{7 x}{7} = \frac{\frac{π}{6}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 x}{7} = \frac{\frac{π}{6}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 x}{7} : x

\frac{7 x}{7}

Divide the numbers:

\frac{7}{7} = 1

= x

Simplify

\frac{\frac{π}{6}}{7} + \frac{2 πn}{7} : \frac{π}{42} + \frac{2 πn}{7}

\frac{\frac{π}{6}}{7} + \frac{2 πn}{7}

\frac{\frac{π}{6}}{7} = \frac{π}{42}

\frac{\frac{π}{6}}{7}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 7}

Multiply the numbers:

6 \cdot 7 = 42

= \frac{π}{42}

= \frac{π}{42} + \frac{2 πn}{7}

x = \frac{π}{42} + \frac{2 πn}{7}

x = \frac{π}{42} + \frac{2 πn}{7}

x = \frac{π}{42} + \frac{2 πn}{7}

Solve

7 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{42} + \frac{2 πn}{7}

7 x = \frac{11 π}{6} + 2 πn

Divide both sides by

7

7 x = \frac{11 π}{6} + 2 πn

Divide both sides by

7

\frac{7 x}{7} = \frac{\frac{11 π}{6}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 x}{7} = \frac{\frac{11 π}{6}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 x}{7} : x

\frac{7 x}{7}

Divide the numbers:

\frac{7}{7} = 1

= x

Simplify

\frac{\frac{11 π}{6}}{7} + \frac{2 πn}{7} : \frac{11 π}{42} + \frac{2 πn}{7}

\frac{\frac{11 π}{6}}{7} + \frac{2 πn}{7}

\frac{\frac{11 π}{6}}{7} = \frac{11 π}{42}

\frac{\frac{11 π}{6}}{7}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 7}

Multiply the numbers:

6 \cdot 7 = 42

= \frac{11 π}{42}

= \frac{11 π}{42} + \frac{2 πn}{7}

x = \frac{11 π}{42} + \frac{2 πn}{7}

x = \frac{11 π}{42} + \frac{2 πn}{7}

x = \frac{11 π}{42} + \frac{2 πn}{7}

x = \frac{π}{42} + \frac{2 πn}{7}, x = \frac{11 π}{42} + \frac{2 πn}{7}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(7x)-sqrt(3)=0 ?
The general solution for 2cos(7x)-sqrt(3)=0 is x= pi/(42)+(2pin)/7 ,x=(11pi}{42}+\frac{2pin)/7