How many solutions on the interval [0, 2020π] does the equation 1/2 Sin2x + 1 = Sinx + Cosx have?
Answers
we have to find the solutions on the interval [0, 2020π] for the equation 1/2 sin2x + 1 = sinx + cosx have.
solution : 1/2 sin2x + 1 = sinx + cosx
⇒1/2(2sinx cosx) + 1/2- 1/2 + 1 = sinx + cosx
⇒1/2 (2sinx cosx ) + 1/2 (sin²x + cos²x) + 1/2 = (sinx + cosx)
[ we know, 1 = sin²x + cos²x ]
⇒1/2 [sin²x + cos²x + 2sinx cosx ] + 1/2 = (sinx + cosx)
⇒1/2 (sinx + cosx)² + 1/2 = (sinx + cosx)
⇒(sinx + cosx)² - 2(sinx + cosx) + 1 = 0
⇒[(sinx + cosx) - 1]² = 0
⇒sinx + cosx = 1
⇒√2[1/√2 sinx + 1/√2 cosx ] = 1
⇒√2[cosπ/4 sinx + sinπ/4 cosx ] = 1
⇒sin(x + π/4) = 1/√2 = sin π/4
so, in [0, 2π], there are two solutions of given equation, x = 0, and π/2
in [0, 2020π] , number of solutions will be
= 2020π/2π × 2
= 2020
Therefore equation will have 2020 solutions on the interval [0, 2020π]