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

The general solution for 2cos(x)=-sqrt(2),0<= x<2pi is x=(3pi)/4 ,x=(5pi)/4

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2cos(x)=-sqrt(2),0<= x<2pi

2 cos (x) = - 2, 0 \leq x < 2 π

x = \frac{3 π}{4}, x = \frac{5 π}{4}

Degrees

x = 13 5^{\circ}, x = 22 5^{\circ}

Solution steps

2 cos (x) = - 2, 0 \leq x < 2 π

Divide both sides by

2

2 cos (x) = - 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (x) = - \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}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{4}, x = \frac{5 π}{4}

What is the general solution for 2cos(x)=-sqrt(2),0<= x<2pi ?
The general solution for 2cos(x)=-sqrt(2),0<= x<2pi is x=(3pi)/4 ,x=(5pi)/4