What is the general solution for 3cos(x)+5=7,0<= x<= 360 ?

The general solution for 3cos(x)+5=7,0<= x<= 360 is x=0.84106…,x=360-0.84106…

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3cos(x)+5=7,0<= x<= 360

3 cos (x) + 5 = 7, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 0.84106 \dots, x = 36 0^{\circ} - 0.84106 \dots

Radians

x = 0.84106 \dots, x = 2 π - 0.84106 \dots

Solution steps

3 cos (x) + 5 = 7, 0^{\circ} \leq x \leq 36 0^{\circ}

Move

5

to the right side

3 cos (x) + 5 = 7

Subtract

5

from both sides

3 cos (x) + 5 - 5 = 7 - 5

Simplify

3 cos (x) = 2

3 cos (x) = 2

Divide both sides by

3

3 cos (x) = 2

Divide both sides by

3

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x = arccos (\frac{2}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{2}{3}) + 36 0^{\circ} n

x = arccos (\frac{2}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{2}{3}) + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = arccos (\frac{2}{3}), x = 36 0^{\circ} - arccos (\frac{2}{3})

Show solutions in decimal form

x = 0.84106 \dots, x = 36 0^{\circ} - 0.84106 \dots

cot (3 x - 6^{\circ}) = \frac{3}{3}

sec (x) = 4

\frac{182.5}{sin ( 36. 6 ^{\circ} )} = \frac{248.6}{sin ( x )}

1 - cos (3 x) = 2 π

(sin (x - 4)) (sin (x - 1)) = 0

What is the general solution for 3cos(x)+5=7,0<= x<= 360 ?
The general solution for 3cos(x)+5=7,0<= x<= 360 is x=0.84106…,x=360-0.84106…