Solve for θ if .

Question

Solve for θ  if  .  ​

Quizplus · Accepted Answer

To solve for θ, we need to first find the complex roots of the given expression in polar form. We can start by factoring out 11ebf5ef_bbb7_8ef3_95d1_47f0b8b6d780_TBX9354_11, which is a common factor:

11ebf5ef_bbb7_8ef3_95d1_47f0b8b6d780_TBX9354_11(1 + e^(iθ))

Now we can use Euler's formula to express 1 + e^(iθ) in polar form:

1 + e^(iθ) = 2(cos(θ/2) + i sin(θ/2))

Substituting back, we have:

11ebf5ef_bbb7_8ef3_95d1_47f0b8b6d780_TBX9354_11(2(cos(θ/2) + i sin(θ/2)))

Now we can convert this expression to rectangular form:

22e^(iθ/2)(cos(θ/2) + i sin(θ/2))

= 22(cos(θ/2)cos(θ/2) - sin(θ/2)sin(θ/2) + i(cos(θ/2)sin(θ/2) + sin(θ/2)cos(θ/2)))

= 22(cos^2(θ/2) - sin^2(θ/2) + 2i(sin(θ/2)cos(θ/2)))

= 22(cosθ + i sinθ)

Where we have used the double angle formula and the fact that sin(θ/2)cos(θ/2) = 1/2 sinθ.

Therefore, we have shown that the given expression can be written as:

22e^(iθ/2)(cos(θ/2) + i sin(θ/2)) = 22(cosθ + i sinθ)

This expression is equivalent to:

11ebf5ef_bbb7_8ef3_95d1_47f0b8b6d780_TBX9354_11(1 + e^(iθ)) = 22(cosθ + i sinθ)

Since the magnitudes of the two factors on the left-hand side are equal, we must have:

1 + e^(iθ) = 2e^(iθ/2)

Solving for e^(iθ/2) gives:

e^(iθ/2) = (1/2)(1 + sqrt(3)i)

Taking the square of both sides:

e^(iθ) = -sqrt(3) + i

We can now use the inverse tangent function to solve for θ:

θ = arctan(1/sqrt(3)) + π

θ ≈ 120°, 240°

Therefore, the correct answer is E) 120°, 240°.

Solve for θ If . ​

Solve for θ If .