2cos^2 x−3cosx+sinx+1=0 help


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Question Number 209656 by Ismoiljon_008 last updated on 17/Jul/24

$2 {c o s}^{2} x - 3 c o s x + s i n x + 1 = 0$ $h e l p$
	Answered by Berbere last updated on 17/Jul/24
	$c=cos s=sin 2c^2 −3c+s+1=0;c^2 +s^2 =1⇒ s^2 +c^2 =(−2c^2 +3c−1)^2 +c^2 =1 ⇔(4c^4 +9c^2 −12c^3 −6c+5c^2 )=0 c(4c^3 −12c^2 +14c−6)=0 ⇒c(2c^3 −6c^2 +7c−3)=0 c.(2c−2)(c^2 −2c+(3/2))=0⇒c∈{0,1} c=0⇒ sin(x)=−1⇒x=−(π/2)+2kπ c=1⇒sin(x)=0⇒x=2kπ S={2kπ;−(π/2)+2kπ};k∈Z$
	$c = c o s$ $s = s i n$ $2 c^{2} - 3 c + s + 1 = 0; c^{2} + s^{2} = 1 \Rightarrow$ $s^{2} + c^{2} = {(- 2 c^{2} + 3 c - 1)}^{2} + c^{2} = 1$ $\Leftrightarrow (4 c^{4} + 9 c^{2} - 12 c^{3} - 6 c + 5 c^{2}) = 0$ $c (4 c^{3} - 12 c^{2} + 14 c - 6) = 0$ $\Rightarrow c (2 c^{3} - 6 c^{2} + 7 c - 3) = 0$ $c . (2 c - 2) (c^{2} - 2 c + \frac{3}{2}) = 0 \Rightarrow c \in {0, 1}$ $c = 0 \Rightarrow s i n (x) = - 1 \Rightarrow x = - \frac{π}{2} + 2 k π$ $c = 1 \Rightarrow s i n (x) = 0 \Rightarrow x = 2 k π$ $S = {2 k π; - \frac{π}{2} + 2 k π}; k \in Z$
	Commented by Ismoiljon_008 last updated on 17/Jul/24

	$t h a n k y o u v e r y m u c h$
	Answered by lepuissantcedricjunior last updated on 18/Jul/24
	$2cos^2 x−3cosx+sinx+1=0 2t^2 −3t+1+(√(1−t^2 ))=0 =>(−2t^2 +3t−1)=(√(1−t^2 )) =>4t^4 −12t^3 +4t^2 −6t+1+9t^2 =1−t^2 =>4t^4 −12t^3 +14t^2 −6t=0 =>t(4t^3 −12t^2 +14t−6)=0 =>t=0ou 4t^3 −12t^2 +14t−6=0 =>t(t−1)(4t^2 −8t+6)=0 =>4t(t−1)(t^2 −2t+(3/2))=0 =>t=0ou t=1 => { ((cosx=0)),((cosx=1)) :}=> { ( { ((x=(𝛑/2)+2k𝛑)),((x=−(𝛑/2)+2k𝛑)) :}),((x=2k𝛑)) :} S_R ={−(𝛑/2)+2k𝛑;(𝛑/2)+2k𝛑;+2k𝛑/k∈Z}$
	$2 c o \overset{2}{s x} - 3 c o s x + s i n x + 1 = 0$ $2 t^{2} - 3 t + 1 + \sqrt{1 - t^{2}} = 0$ $=> (- 2 t^{2} + 3 t - 1) = \sqrt{1 - t^{2}}$ $=> 4 t^{4} - 12 t^{3} + 4 t^{2} - 6 t + 1 + 9 t^{2} = 1 - t^{2}$ $=> 4 t^{4} - 12 t^{3} + 14 t^{2} - 6 t = 0$ $=> t (4 t^{3} - 12 t^{2} + 14 t - 6) = 0$ $=> t = 0 o u 4 t^{3} - 12 t^{2} + 14 t - 6 = 0$ $=> t (t - 1) (4 t^{2} - 8 t + 6) = 0$ $=> 4 t (t - 1) (t^{2} - 2 t + \frac{3}{2}) = 0$ $=> t = 0 o u t = 1$ $=> {\begin{cases} c o s x = 0 \\ c o s x = 1 \end{cases} => {\begin{cases} {\begin{cases} x = \frac{π}{2} + 2 k π \\ x = - \frac{π}{2} + 2 k π \end{cases} \\ x = 2 k π \end{cases}$ $S_{R} = {- \frac{π}{2} + 2 k π; \frac{π}{2} + 2 k π; + 2 k π / k \in Z}$
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