Multiple choice

If $\sin\theta+7\cos \theta=5$, then $\tan(\theta/2)$ is a root of the equation :

  1. $x^{2}-6x+1=0$
  2. $6x^{2}-x-1=0$
  3. $6x^{2}+x+1=0$
  4. $x^{2}-x+6=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let t = tan(theta/2). sin(theta) = 2t/(1+t^2), cos(theta) = (1-t^2)/(1+t^2). Substitute: 2t/(1+t^2) + 7(1-t^2)/(1+t^2) = 5. 2t + 7 - 7t^2 = 5 + 5t^2. 12t^2 - 2t - 2 = 0. 6t^2 - t - 1 = 0.

AI explanation

Using the universal trigonometric substitution, let tan(theta/2) = x, which converts sin(theta) into 2x/(1 + x^2) and cos(theta) into (1 - x^2)/(1 + x^2). Substituting these into the equation sin(theta) + 7cos(theta) = 5 gives 2x + 7(1 - x^2) = 5(1 + x^2). Rearranging the terms yields 12x^2 - 2x - 2 = 0, which simplifies by dividing by 2 to 6x^2 - x - 1 = 0. Thus, tan(theta/2) is a root of this quadratic equation.

Keep practicing — related questions