Trig Q: express cos(2x) - sqrt3 sin(2x) in the form of Rcos(2x+alpha)

[devgru11]

1) express cos(2x) - sqrt3 sin(2x) in the form of Rcos(2x+alpha)

Answer: 2cos(2x+pi/3). can anyone confirm if this is the correct answer?

1b) use the above result to find the exact coordinates of the intercepts of the graph of cos(2x) - sqrt3 sin(2x) -1 , x E [-pi,pi]

my answer is: cos(2x+pi/3) = 1/2 = (pi/3)

so x = -pi, -pi/3, 0, 2pi/3, pi.

but why does the answer say x = -2.06669, -1.07490, 1.07490, 2.06669

 

[Deleted member 4993]

1) express cos(2x) - sqrt3 sin(2x) in the form of Rcos(2x+alpha)

Answer: 2cos(2x+pi/3). can anyone confirm if this is the correct answer? .... expand 2cos(2x+π/3) - what do you get?

1b) use the above result to find the exact coordinates of the intercepts of the graph of cos(2x) - sqrt3 sin(2x) -1 , x E [-pi,pi]

my answer is: cos(2x+pi/3) = 1/2 = cos(±pi/3 ± 2 * k * π)

so x = -pi, -pi/3, 0, 2pi/3, pi.

but why does the answer say x = -2.06669, -1.07490, 1.07490, 2.06669

[2] x E [-pi,pi] excludes ±π from the domain.

The answer (for the given domain) should be x = - π/3, 0, 2π/3 ..... these are the exact co-ordinates.

You can check your (and books) answers by replacing 'x' by corresponding values and evaluating the given expression. Observe whether the "equality" holds.