What is the general solution for 3cos(x)+5=7,0<= x<= 2pi ?

The general solution for 3cos(x)+5=7,0<= x<= 2pi is x=0.84106…,x=2pi-0.84106…

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3cos(x)+5=7,0<= x<= 2pi

3 cos (x) + 5 = 7, 0 \leq x \leq 2 π

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

Solution steps

3 cos (x) + 5 = 7, 0 \leq x \leq 2 π

Move

5

to the right side

3 cos (x) = 2

Divide both sides by

3

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

Apply trig inverse properties

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = arccos (\frac{2}{3}), x = 2 π - arccos (\frac{2}{3})

Show solutions in decimal form

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

9 sin (\frac{x}{2}) + 9 cos (x) = 0

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

10 sin (θ) = 4 cos^{2} (θ) + 2

cos (x) = - \frac{1}{3}

6 - 24 cos^{2} (t) = 0

What is the general solution for 3cos(x)+5=7,0<= x<= 2pi ?
The general solution for 3cos(x)+5=7,0<= x<= 2pi is x=0.84106…,x=2pi-0.84106…