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

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

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2cos(A)=-sqrt(3)

2 cos (A) = - 3

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

Degrees

A = 15 0^{\circ} + 36 0^{\circ} n, A = 21 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (A) = - 3

Divide both sides by

2

2 cos (A) = - 3

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (A) = - \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}

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

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

4 sin (x) cos (x) = cos (x)

cos (x) = 2 sin (x) cos (x)

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

tan^{2} (4 5^{\circ} + \frac{x}{3}) = 0.58

1 = 4 cos^{2} (x) - 4 cos (x) + 2

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