Solve 3cos(2x)=2 for the smallest three positive solutions.

Question

Solve 3cos(2x)=2 for the smallest three positive solutions.

Give your answers accurate to at least two decimal places, as a list separated by commas

Answer 1

3cos(2x) = 2

cos(2x) = 2/3

2x = Cos^-1(2/3)

2x = 48.19019757°(first quadrant), 360° - 48.19019757°(fourth quadrant), or 48.19019757° + 360°(cosine repeats every 360°)

x = 24.10°, 155.90°, or 204.10°,

Answer 2

3 cos(2x) = 2

cos(2x) = 2/3

2x = arccos(2/3)

= 48.189685104221401934142083269422

so x = 24.094842552110700967071041634711 degrees

is the base angle

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Next,

3 [2 cos^2– 1] = 2

6 cos^2 - 3 = 2

6 cos^2 = 5

cos^2 = 5/6

cos = +or-sqrt(5/6)

cos = +or- sqrt(30)/6 <--- rationializes the denominator

x = arccos(sqrt(30)/6) = 24.094842552110700967071041634711

cosine is positive in quadrants 1 and 4

Q4, the angle is 360-24.094842552110700967071041634711 =

335.90515744788929903292895836529

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sorry, the 3rd positive solution is 360+ 24.094842552110700967071041634711 =

384.094842552110700967071041634711