What is the general solution for 40cos((2pi)/3 t)-71=-80 ?

The general solution for 40cos((2pi)/3 t)-71=-80 is t=(3*1.79773…)/(2pi)+3n,t=-(3*1.79773…)/(2pi)+3n

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40cos((2pi)/3 t)-71=-80

40 cos (\frac{2 π}{3} t) - 71 = - 80

t = \frac{3 \cdot 1.79773 \dots}{2 π} + 3 n, t = - \frac{3 \cdot 1.79773 \dots}{2 π} + 3 n

Solution steps

40 cos (\frac{2 π}{3} t) - 71 = - 80

Move

71

to the right side

40 cos (\frac{2 π}{3} t) = - 9

Divide both sides by

40

cos (\frac{2 π}{3} t) = - \frac{9}{40}

Apply trig inverse properties

\frac{2 π}{3} t = arccos (- \frac{9}{40}) + 2 πn, \frac{2 π}{3} t = - arccos (- \frac{9}{40}) + 2 πn

Solve

\frac{2 π}{3} t = arccos (- \frac{9}{40}) + 2 πn : t = \frac{3 arccos ( - \frac{9}{40} )}{2 π} + 3 n

Solve

\frac{2 π}{3} t = - arccos (- \frac{9}{40}) + 2 πn : t = - \frac{3 arccos ( - \frac{9}{40} )}{2 π} + 3 n

t = \frac{3 arccos ( - \frac{9}{40} )}{2 π} + 3 n, t = - \frac{3 arccos ( - \frac{9}{40} )}{2 π} + 3 n

Show solutions in decimal form

t = \frac{3 \cdot 1.79773 \dots}{2 π} + 3 n, t = - \frac{3 \cdot 1.79773 \dots}{2 π} + 3 n

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

tan (x) = 2 sin^{2} (x)

0 = 4 sin (2 x)

\frac{sin ( 112. 8 ^{\circ} )}{16.1} = \frac{sin ( x )}{12.2}

(sin (x) + cos (x))^{2} = \frac{3}{2}

What is the general solution for 40cos((2pi)/3 t)-71=-80 ?
The general solution for 40cos((2pi)/3 t)-71=-80 is t=(3*1.79773…)/(2pi)+3n,t=-(3*1.79773…)/(2pi)+3n