Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 157080 by amin96 last updated on 19/Oct/21

Commented by cortano last updated on 20/Oct/21
$⇒1+((1−cos 2x)/2)=((3sin 2x)/2) ⇒3−cos 2x=3sin 2x ⇒3−cos 2x=3(√(1−cos^2 (2x))) ⇒9+cos^2 (2x)−6cos (2x)=9−9cos^2 (2x) ⇒10cos^2 (2x)−6cos (2x)=0 ⇒cos 2x(5cos (2x)−3)=0 { ((cos 2x=0⇒x=±(π/4)+nπ)),((cos 2x=(3/5)⇒tan (2x)=(4/3))) :} ⇒((2tan x)/(1−tan^2 x))=(4/3) ⇒6tan x=4−4tan^2 x ⇒2tan^2 x+3tan x−2=0 ⇒(2tan x−1)(tan x+2)=0 Σ tan (x_i )= −(3/2)$
$\Rightarrow 1 + \frac{1 - \cos 2 x}{2} = \frac{3 \sin 2 x}{2}$ $\Rightarrow 3 - \cos 2 x = 3 \sin 2 x$ $\Rightarrow 3 - \cos 2 x = 3 \sqrt{1 - \cos^{2} (2 x)}$ $\Rightarrow 9 + \cos^{2} (2 x) - 6 \cos (2 x) = 9 - 9 \cos^{2} (2 x)$ $\Rightarrow 10 \cos^{2} (2 x) - 6 \cos (2 x) = 0$ $\Rightarrow \cos 2 x (5 \cos (2 x) - 3) = 0$ ${\begin{cases} \cos 2 x = 0 \Rightarrow x = \pm \frac{π}{4} + n π \\ \cos 2 x = \frac{3}{5} \Rightarrow \tan (2 x) = \frac{4}{3} \end{cases}$ $\Rightarrow \frac{2 \tan x}{1 - \tan^{2} x} = \frac{4}{3}$ $\Rightarrow 6 \tan x = 4 - 4 \tan^{2} x$ $\Rightarrow 2 \tan^{2} x + 3 \tan x - 2 = 0$ $\Rightarrow (2 \tan x - 1) (\tan x + 2) = 0$ $Σ \tan (x_{i}) = - \frac{3}{2}$
Commented by mr W last updated on 19/Oct/21

$\tan x = \frac{1}{2} o r 1$ $\Rightarrow Σ \tan x = \frac{3}{2}$
	Answered by mr W last updated on 19/Oct/21

	$1 + \frac{1}{2} (1 - \cos 2 x) = \frac{3}{2} \sin 2 x$ $3 \sin 2 x + \cos 2 x = 3$ $l e t t = \tan x$ $3 \times \frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}} = 3$ $2 t^{2} - 3 t + 1 = 0$ $Σ t = Σ \tan x = \frac{3}{2}$
	Answered by amin96 last updated on 19/Oct/21
	$(1/(cos^2 (x)))+((sin^2 (x))/(cos^2 (x)))=3((sin(x)cos(x))/(cos^2 (x))) 1+tg^2 (x)+t^2 (x)=3tg(x) 2tg^2 (x)−3tg(x)+1=0 2a^2 −3a+1=0 { ((a_1 =tg(x)_1 =(1/2))),((a_2 =tg(x)_2 =1)) :} a_1 +a_2 =1,5 I think a solution like this would be easier.$
	$\frac{1}{\cos^{2} (x)} + \frac{\sin^{2} (x)}{\cos^{2} (x)} = 3 \frac{\sin (x) \cos (x)}{\cos^{2} (x)}$ $1 + {tg}^{2} (x) + t^{2} (x) = 3 tg (x)$ $2 {tg}^{2} (x) - 3 tg (x) + 1 = 0$ $2 a^{2} - 3 a + 1 = 0 {\begin{cases} a_{1} = tg {(x)}_{1} = \frac{1}{2} \\ a_{2} = tg {(x)}_{2} = 1 \end{cases}$ $a_{1} + a_{2} = 1, 5$ I think a solution like this would be easier.
	Commented by amin96 last updated on 19/Oct/21

	just occurred to me)
	Answered by lyubita last updated on 19/Oct/21

	$\sin^{2} x + \cos^{2} x + \sin^{2} x = 3 \sin x \cos x$ $2 \sin^{2} x - 3 \sin x \cos x + \cos^{2} x = 0$ $(2 \sin x - \cos x) (\sin x - \cos x) = 0$ $2 \sin x = \cos x \sin x = \cos x$ $\tan x = \frac{1}{2} \tan x = 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum