cos-x-1-sin-x-2cos-x-cos-x-sin-x-in-5pi-7pi-2-

Question Number 81062 by john santu last updated on 09/Feb/20

\cos x \sqrt{1 + \sin x - 2 \cos x} = \cos x - \sin x

i n [- 5 π, - \frac{7 π}{2}]

Commented by MJS last updated on 09/Feb/20

sorry no time today

(1) let x = 2 \arctan t

\Rightarrow \sin x = \frac{2 t}{t^{2} + 1} \land \cos x = - \frac{t^{2} - 1}{t^{2} + 1}

(2) check for which values the \sqrt{\dots} \in R

(3) solve for t

[squaring \Rightarrow not all solutions are useable!]

(4) t = \tan \frac{x}{2} \Leftrightarrow x = 2 n π + 2 \arctan t

find the solution in the required interval

Commented by jagoll last updated on 09/Feb/20

h e l l o m i s t e r . h o w a r e y o u ?

Commented by john santu last updated on 09/Feb/20

t h a n k y o u s i r

Commented by MJS last updated on 09/Feb/20

\dots this leads to a polynome of 6^{th} degree :

t^{6} - t^{5} - 5 t^{4} - 2 t^{3} + t^{2} + 3 t - 1 = 0

which can be factorized \dots

(t^{2} + t - 1) (t^{2} - (1 + \sqrt{5}) t + 1) (t^{2} - (1 - \sqrt{5}) t + 1) = 0

\dots and solved

Commented by john santu last updated on 09/Feb/20

m y w a y s i r

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

(2) 1 + \sin x - 2 \cos x ⩾ 0

(3) \cos x (\cos x - \sin x) > 0

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

(i) \sin x = 2 \cos x \Rightarrow \tan x = 2

(i i) \cos^{2} x - \sin x = 0

\sin^{2} x + \sin x - 1 = 0 \Rightarrow \sin x = \frac{- 1 + \sqrt{5}}{2}

x = \sin^{- 1} (\frac{- 1 + \sqrt{5}}{2}) + 2 π n, n \in Z

x = π - \sin^{- 1} (\frac{- 1 + \sqrt{5}}{2}) + 2 π n, n \in Z

(3) \cos x (\cos x - \sin x) > 0

\cos x = 0 \Rightarrow x = \frac{π}{2} + π n, n \in Z

\cos x = \sin x \Rightarrow x = \frac{π}{4} + π k, k \in Z