What is the general solution for 3cos(z+1.2)=2 ?

The general solution for 3cos(z+1.2)=2 is z=0.84106…+2pin-1.2,z=2pi-0.84106…+2pin-1.2

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3cos(z+1.2)=2

3 cos (z + 1.2) = 2

z = 0.84106 \dots + 2 πn - 1.2, z = 2 π - 0.84106 \dots + 2 πn - 1.2

Degrees

z = - 20.56525 \dots^{\circ} + 36 0^{\circ} n, z = 243.05537 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (z + 1.2) = 2

Divide both sides by

3

3 cos (z + 1.2) = 2

Divide both sides by

3

\frac{3 cos ( z + 1.2 )}{3} = \frac{2}{3}

Simplify

cos (z + 1.2) = \frac{2}{3}

cos (z + 1.2) = \frac{2}{3}

Apply trig inverse properties

cos (z + 1.2) = \frac{2}{3}

General solutions for

cos (z + 1.2) = \frac{2}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

z + 1.2 = arccos (\frac{2}{3}) + 2 πn, z + 1.2 = 2 π - arccos (\frac{2}{3}) + 2 πn

z + 1.2 = arccos (\frac{2}{3}) + 2 πn, z + 1.2 = 2 π - arccos (\frac{2}{3}) + 2 πn

Solve

z + 1.2 = arccos (\frac{2}{3}) + 2 πn : z = arccos (\frac{2}{3}) + 2 πn - 1.2

z + 1.2 = arccos (\frac{2}{3}) + 2 πn

Move

1.2

to the right side

z + 1.2 = arccos (\frac{2}{3}) + 2 πn

Subtract

1.2

from both sides

z + 1.2 - 1.2 = arccos (\frac{2}{3}) + 2 πn - 1.2

Simplify

z = arccos (\frac{2}{3}) + 2 πn - 1.2

z = arccos (\frac{2}{3}) + 2 πn - 1.2

Solve

z + 1.2 = 2 π - arccos (\frac{2}{3}) + 2 πn : z = 2 π - arccos (\frac{2}{3}) + 2 πn - 1.2

z + 1.2 = 2 π - arccos (\frac{2}{3}) + 2 πn

Move

1.2

to the right side

z + 1.2 = 2 π - arccos (\frac{2}{3}) + 2 πn

Subtract

1.2

from both sides

z + 1.2 - 1.2 = 2 π - arccos (\frac{2}{3}) + 2 πn - 1.2

Simplify

z = 2 π - arccos (\frac{2}{3}) + 2 πn - 1.2

z = 2 π - arccos (\frac{2}{3}) + 2 πn - 1.2

z = arccos (\frac{2}{3}) + 2 πn - 1.2, z = 2 π - arccos (\frac{2}{3}) + 2 πn - 1.2

Show solutions in decimal form

z = 0.84106 \dots + 2 πn - 1.2, z = 2 π - 0.84106 \dots + 2 πn - 1.2

\frac{1}{2} = \frac{3}{2} \cdot cos (x)

cos (\frac{x}{2}) = - 12

4 sin (β) + 2 = 3 + 2 sin (β)

3 sin^{2} (x) = cos (x)

cos (3 x) = 0, 0 \leq x \leq 2 π

What is the general solution for 3cos(z+1.2)=2 ?
The general solution for 3cos(z+1.2)=2 is z=0.84106…+2pin-1.2,z=2pi-0.84106…+2pin-1.2