Question

Let m be the number of solutions of sin(2x) +cos(2x) + cosx + 1 = 0 in 0 < x < π/2 and n = sin[tan^-1(tan(7π/6)) + cos^-1(cos(7π/3))] then find the value of (m² + n² + m + n + 4)

Answer 1

We have

sin(2x) +cos(2x) + cosx + 1 = 0

sin2x + (1 + cos2x) + cosx = 0

Each term of the above equation is positive in (0, π/2).

So it has no solution

Thus, m = 0

Also, n = sin[tan^-1(tan(7π/6)) + cos^-1(cos(7π/3))]

= sin(π/6 + π/3) = sin(π/2) = 1

Hence, the value of

(m² + n² + m + n + 4)

= 0 + 1 + 0 + 1 + 4

= 6