Solve-for-real-numbers-3sinx-4-y-cosx-y-2-9-

Question Number 179449 by Shrinava last updated on 29/Oct/22

Solve for real numbers :

3 sinx + 4 (y + cosx) = y^{2} + 9

Answered by mr W last updated on 29/Oct/22

3 \sin x + 4 \cos x - 5 = y^{2} - 4 y + 4

5 [\sin (x + α) - 1] = {(y - 2)}^{2}

s i n c e \sin (x + α) ⩽ 1, L H S ⩽ 0 .

b u t R H S ⩾ 0, t h e r e f o r e

L H S = R H S = 0

\Rightarrow \sin (x + α) = 1 \Rightarrow x = 2 k π + \frac{π}{2} - \tan^{- 1} \frac{4}{3}

\Rightarrow y = 2

Commented by Shrinava last updated on 01/Nov/22

cool dear professor thank you

Answered by manxsol last updated on 29/Oct/22

3 s i n x + 4 c o s x = y^{2} - 4 y + 9

\frac{3}{5} s i n x + \frac{4}{5} c o s x = \frac{y^{2} - 4 y + 9}{5}

s i n (x + θ) = \frac{y^{2} - 4 y + 9}{5}

θ = a r s e n (3 / 5)

- 1 ⩽ \frac{y^{2} - 4 y + 9}{5} ⩽ 1

- 5 ⩽ y^{2} - 4 y + 9 \hat{Λ} y^{2} - 4 y + 9 ⩽ 5

0 ⩽ y^{2} - 4 y + 14 Δ = 16 - 56

Δ < 0 \Rightarrow y^{2} - 4 y + 14 ⟩ 0 \forall y >

y^{2} - 4 y + 9 ⩽ 5

y^{2} - 4 y + 4 ⩽ 0

{(y - 2)}^{2} ⩽ 0

s o l u t i o n y = 2

s i n (x + θ) = \frac{2^{2} - 4 (2) + 9}{5} = 1

x + θ = \frac{π}{2} + 2 k π

x = (\frac{π}{2} - a r c s i n (\frac{3}{5})) + 2 k π

Commented by Shrinava last updated on 01/Nov/22

cool dear professor thank you