cos-2-x-tan-2-x-3-2-x-R-

Question Number 144379 by imjagoll last updated on 25/Jun/21

\cos^{2} x + \tan^{2} x = \frac{3}{2}

x ϵ R

Answered by liberty last updated on 25/Jun/21

\cos^{2} x + \tan^{2} x = \frac{3}{2}

\cos^{2} x + \sec^{2} x - \frac{5}{2} = 0

set \cos^{2} x = z; 0 < z ⩽ 1

{2 z}^{2} - 5 z + 2 = 0

(2 z - 1) (z - 2) = 0

z = \frac{1}{2} \Rightarrow \cos x = \pm \frac{\sqrt{2}}{2}

when \cos x = - \frac{\sqrt{2}}{2}; x = 2 n π \pm \frac{3 π}{4}

when \cos x = \frac{\sqrt{2}}{2}; x = 2 n π \pm \frac{π}{4}

Answered by yoba last updated on 25/Jun/21

\cos^{2} x + \tan^{2} x = \cos^{2} x + 1 + \tan^{2} x - 1 = \cos^{2} x + \frac{1}{\cos^{2} x} - 1 = \frac{3}{2}

∙ contraintes sur x : cosx \neq 0 \Rightarrow x \in R ∖ {- \frac{π}{2} [k π], k \in Z}

\Rightarrow \cos^{2} x + \frac{1}{\cos^{2} x} - \frac{5}{2} = 0 \Rightarrow \cos^{4} x - \frac{5}{2} \cos^{2} x + 1 = 0

\Rightarrow 2 \cos^{4} x - 5 \cos^{2} x + 2 = 0 posons a = \cos^{2} x

\Rightarrow 2 a^{2} - 5 a + 2 = 0 \Rightarrow △ = 25 - 4 \times 4 = 9 = 3^{2}

\Rightarrow a = \frac{1}{2} ou a = 2

\Rightarrow \cos^{2} x = \frac{1}{2} ou \cos^{2} x = 2 (impossible)

\Rightarrow cosx = \frac{\sqrt{2}}{2} ou cosx = - \frac{\sqrt{2}}{2}

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