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

The general solution for 2cos(A)=-sqrt(2) is A=(3pi)/4+2pin,A=(5pi)/4+2pin

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2cos(A)=-sqrt(2)

2 cos (A) = - 2

A = \frac{3 π}{4} + 2 πn, A = \frac{5 π}{4} + 2 πn

Degrees

A = 13 5^{\circ} + 36 0^{\circ} n, A = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (A) = - 2

Divide both sides by

2

2 cos (A) = - 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (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}

A = \frac{3 π}{4} + 2 πn, A = \frac{5 π}{4} + 2 πn

A = \frac{3 π}{4} + 2 πn, A = \frac{5 π}{4} + 2 πn

6 sin (x) + 5 cos (x) = 0

2 cos^{2} (x) - 5 cos (x) + 3 = 0

2 sin^{3} (x) = 0

csc (x) = - 2

6 cos (A) + 18 = 0

What is the general solution for 2cos(A)=-sqrt(2) ?
The general solution for 2cos(A)=-sqrt(2) is A=(3pi)/4+2pin,A=(5pi)/4+2pin