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

The general solution for cos(x)=-3/(2sqrt(3)) is x=(5pi)/6+2pin,x=(7pi)/6+2pin

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cos(x)=-3/(2sqrt(3))

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

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

Solution steps

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

Simplify

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

- \frac{3}{2 3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3}{2 \cdot 3 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= \frac{3 ^{1 - \frac{1}{2}}}{2}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{3 ^{\frac{1}{2}}}{2}

Apply radical rule:

3^{\frac{1}{2}} = 3

= - \frac{3}{2}

General solutions for

cos (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}

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

sec^{2} (x) - 6 sec (x) = 16

cos^{2} (θ) + sin^{2} (θ) = 4

6 arccos (x) - 3 π = 3 arccos (x)

3 sin (2 x) = sin (x)

cos (b) = \frac{12}{37}

What is the general solution for cos(x)=-3/(2sqrt(3)) ?
The general solution for cos(x)=-3/(2sqrt(3)) is x=(5pi)/6+2pin,x=(7pi)/6+2pin