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𝑺𝒐𝒍𝒗𝒆,π’ˆπ’Šπ’—π’†π’ 𝒕𝒉𝒂𝒕 πŸŽβ‰€π’™<πŸπ…

 

a) π’”π’Šπ’^2(𝒙)βˆ’π’”π’Šπ’(𝒙)=𝟐

 

b) 𝒄𝒐𝒔^2(𝒙)+𝒄𝒐𝒔(𝒙)=π’”π’Šπ’^2(𝒙)

 Dec 17, 2016

Best Answer 

 #2
avatar+9665 
+5

\(\sin^2 x - \sin x = 2\\ \sin^2 x - \sin x - 2 = 0 \\\text{Let a = sin x}\\ a^2 - a - 2 = 0\\ (a-2)(a+1) = 0\\ (\sin x - 2)(\sin x + 1) = 0\\ \sin x = 2\text{(rejected) or }\sin x = -1\\ \sin x = -1\\ x = \dfrac{3\pi}{2} \text{rad}\)

 

\(\cos^2 x + \cos x = \sin^2x\\ \cos^2x + \cos x = 1 - \cos^2x\\ 2\cos^2x + \cos x - 1 = 0\\ \text{Let a = cos x this time.}\\ 2a^2 + a - 1 = 0\\ (2a - 1)(a + 1) = 0\\ \cos x = \dfrac{1}{2} \text{ or }\cos x = -1\\ x = \dfrac{\pi}{3}\text{ rad} \text{ or }x = \pi \text{ rad}\)

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 Dec 17, 2016
 #1
avatar+37084 
0

Not sure how to prove it, but just by looking at it 'a' is 3/2 pi

 Dec 17, 2016
 #2
avatar+9665 
+5
Best Answer

\(\sin^2 x - \sin x = 2\\ \sin^2 x - \sin x - 2 = 0 \\\text{Let a = sin x}\\ a^2 - a - 2 = 0\\ (a-2)(a+1) = 0\\ (\sin x - 2)(\sin x + 1) = 0\\ \sin x = 2\text{(rejected) or }\sin x = -1\\ \sin x = -1\\ x = \dfrac{3\pi}{2} \text{rad}\)

 

\(\cos^2 x + \cos x = \sin^2x\\ \cos^2x + \cos x = 1 - \cos^2x\\ 2\cos^2x + \cos x - 1 = 0\\ \text{Let a = cos x this time.}\\ 2a^2 + a - 1 = 0\\ (2a - 1)(a + 1) = 0\\ \cos x = \dfrac{1}{2} \text{ or }\cos x = -1\\ x = \dfrac{\pi}{3}\text{ rad} \text{ or }x = \pi \text{ rad}\)

MaxWong Dec 17, 2016
 #3
avatar+37084 
+5

Nicely done, Max !  Thanx !

 Dec 17, 2016

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