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Solve the equation 2\[0.5ex] theta - 3 theta − 2 = 0 for 0^\[0.5ex] theta 360^\[0.5ex] using the t-formula.

Mathematics

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here's how to solve the equation 2cos⁡θ−3sin⁡θ−2=02\cos \theta - 3\sin \theta - 2 = 0 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ using the t-formula:

  1. Apply the t-formula substitutions.

    Let t=tan⁡(θ2)t = \tan\left(\frac{\theta}{2}\right). The t-formula states that sin⁡θ=2t1+t2\sin \theta = \frac{2t}{1+t^2} and cos⁡θ=1−t21+t2\cos \theta = \frac{1-t^2}{1+t^2}.
    Substitute these into the given equation:

    2(1−t21+t2)−3(2t1+t2)−2=02\left(\frac{1-t^2}{1+t^2}\right) - 3\left(\frac{2t}{1+t^2}\right) - 2 = 0
  2. Clear the denominator and simplify the equation.

    Multiply the entire equation by (1+t2)(1+t^2) to eliminate the denominators:

    2(1−t2)−3(2t)−2(1+t2)=02(1-t^2) - 3(2t) - 2(1+t^2) = 0

    Expand the terms:

    2−2t2−6t−2−2t2=02 - 2t^2 - 6t - 2 - 2t^2 = 0

    Combine like terms:

    −4t2−6t=0-4t^2 - 6t = 0

    Divide by −2-2 to simplify:

    2t2+3t=02t^2 + 3t = 0
  3. Solve the quadratic equation for tt.

    Factor out tt:

    t(2t+3)=0t(2t + 3) = 0

    This gives two possible values for tt:

    t=0or2t+3=0  ⟹  t=−32t = 0 \quad \text{or} \quad 2t + 3 = 0 \implies t = -\frac{3}{2}
  4. Solve for θ\theta using t=tan⁡(θ2)t = \tan\left(\frac{\theta}{2}\right).

    Case 1: t=0t = 0

    tan⁡(θ2)=0\tan\left(\frac{\theta}{2}\right) = 0

    The general solution for tan⁡x=0\tan x = 0 is $x = n \cdot 180

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