Multiple choice

Consider the cubic equation ${ x }^{ 3 }-\left( 1+\cos { \theta } +\sin { \theta } \right) { x }^{ 2 }+\left( \cos { \theta } \sin { \theta } +\cos { \theta } +\sin { \theta } \right) x-\sin { \theta } \cos { \theta } =0$ whose roots are ${ x }{ 1 },{ x }{ 2 },{ x }_{ 3 }$ The number of values of $\displaystyle \theta $ in $\displaystyle \left [ 0,2\pi \right ]$ for which at least two roots are equal

  1. $3$
  2. $4$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The cubic equation is (x-1)(x-sin theta)(x-cos theta) = 0. Roots are 1, sin theta, cos theta. At least two roots are equal if: 1 = sin theta (theta = pi/2), 1 = cos theta (theta = 0, 2pi), or sin theta = cos theta (theta = pi/4, 5pi/4). Total values: pi/2, 0, 2pi, pi/4, 5pi/4. That is 5 values.