What is the general solution for 3cos((2pit)/3)+10=11 ?

The general solution for 3cos((2pit)/3)+10=11 is t=(3*1.23095…)/(2pi)+3n,t=3-(3*1.23095…)/(2pi)+3n

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3cos((2pit)/3)+10=11

3 cos (\frac{2 π t}{3}) + 10 = 11

t = \frac{3 \cdot 1.23095 \dots}{2 π} + 3 n, t = 3 - \frac{3 \cdot 1.23095 \dots}{2 π} + 3 n

Solution steps

3 cos (\frac{2 π t}{3}) + 10 = 11

Move

10

to the right side

3 cos (\frac{2 π t}{3}) = 1

Divide both sides by

3

cos (\frac{2 π t}{3}) = \frac{1}{3}

Apply trig inverse properties

\frac{2 π t}{3} = arccos (\frac{1}{3}) + 2 πn, \frac{2 π t}{3} = 2 π - arccos (\frac{1}{3}) + 2 πn

Solve

\frac{2 π t}{3} = arccos (\frac{1}{3}) + 2 πn : t = \frac{3 arccos ( \frac{1}{3} )}{2 π} + 3 n

Solve

\frac{2 π t}{3} = 2 π - arccos (\frac{1}{3}) + 2 πn : t = 3 - \frac{3 arccos ( \frac{1}{3} )}{2 π} + 3 n

t = \frac{3 arccos ( \frac{1}{3} )}{2 π} + 3 n, t = 3 - \frac{3 arccos ( \frac{1}{3} )}{2 π} + 3 n

Show solutions in decimal form

t = \frac{3 \cdot 1.23095 \dots}{2 π} + 3 n, t = 3 - \frac{3 \cdot 1.23095 \dots}{2 π} + 3 n

cos (θ) = \frac{6}{6}, sin (θ)

9 sec^{2} (x) - 9 = 0

4 tan (x) - 7 = 3 tan (x) - 6

- 6 cos (2 θ) + 2 cos (θ) + 2 = 3 cos (θ)

23 cos (2 θ) = - 3

What is the general solution for 3cos((2pit)/3)+10=11 ?
The general solution for 3cos((2pit)/3)+10=11 is t=(3*1.23095…)/(2pi)+3n,t=3-(3*1.23095…)/(2pi)+3n