Find-a-b-c-for-equation-acos-2x-bsin-2-x-c-0-is-satisfied-by-every-x-

Question Number 76622 by Sathvik last updated on 28/Dec/19

Find (a . b . c) for equation acos 2 x + bsin^{2} x + c = 0 is satisfied by every x

Answered by MJS last updated on 28/Dec/19

a \cos 2 x + b \sin^{2} x + c = 0

a \cos 2 x + b \times \frac{1}{2} (1 - \cos 2 x) + c = 0

(a - \frac{b}{2}) \cos 2 x + \frac{b}{2} + c = 0

p \cos 2 x + q = 0 \forall x \in R \Rightarrow p = q = 0

a - \frac{b}{2} = \frac{b}{2} + c \Rightarrow a = b + c

a - \frac{b}{2} = 0 \to \frac{b}{2} + c = 0 \Rightarrow b = - 2 c \Rightarrow a = - c

c = - t \in R

b = 2 t

a = t

\Rightarrow t \cos 2 x + 2 t \sin^{2} x - t = 0

t (\cos 2 x + {2 \sin}^{2} x - 1) = 0

but \cos 2 x + {2 \sin}^{2} x - 1 = 0

\Rightarrow 0 t = 0 true for all t, x \in R

Answered by john santu last updated on 29/Dec/19

a \cos 2 x + b (\frac{1}{2} - \frac{1}{2} \cos 2 x) = - c

(a - \frac{1}{2} b) \cos 2 x = - c - \frac{1}{2} b

\cos 2 x = \frac{- 2 c - b}{2 a - b} \to - 1 ⩽ \frac{- 2 c - b}{2 a - b} ⩽ 1