What is the general solution for 5cos(x)+2=0,0<= x<= 2pi ?

The general solution for 5cos(x)+2=0,0<= x<= 2pi is x=1.98231…,x=-1.98231…+2pi

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5cos(x)+2=0,0<= x<= 2pi

5 cos (x) + 2 = 0, 0 \leq x \leq 2 π

x = 1.98231 \dots, x = - 1.98231 \dots + 2 π

Solution steps

5 cos (x) + 2 = 0, 0 \leq x \leq 2 π

Move

2

to the right side

5 cos (x) = - 2

Divide both sides by

5

cos (x) = - \frac{2}{5}

Apply trig inverse properties

x = arccos (- \frac{2}{5}) + 2 πn, x = - arccos (- \frac{2}{5}) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = arccos (- \frac{2}{5}), x = - arccos (- \frac{2}{5}) + 2 π

Show solutions in decimal form

x = 1.98231 \dots, x = - 1.98231 \dots + 2 π

2 sin (x) cos (x) - sin (2 x) cos (2 x) = 0

cos (x) = \frac{21}{29}

\frac{4}{sin ( 104.4 8 ^{\circ} )} = \frac{2}{sin ( a )}

sin (x + \frac{3 π}{2}) - cos (x + \frac{π}{2}) = 1

csc (x) - cot (x) = 1

What is the general solution for 5cos(x)+2=0,0<= x<= 2pi ?
The general solution for 5cos(x)+2=0,0<= x<= 2pi is x=1.98231…,x=-1.98231…+2pi