Corrigés - inéquation trigonométrique - TS2

Corrigé 1

Dans $]-\pi\ ;\ \pi]$

1. $x\in\left\lbrace -\dfrac{3\pi}{4}\;,-\dfrac{5\pi}{12}\;,\dfrac{\pi}{4}\;,\dfrac{7\pi}{12}\right\rbrace$

2. $x\in\left\lbrace -\dfrac{2\pi}{3}\;,-\dfrac{\pi}{2}\;,0\;,\dfrac{\pi}{6}\;,\dfrac{2\pi}{3}\;,\dfrac{5\pi}{6}\right\rbrace$

3. $x\in\left\lbrace 0\;,\dfrac{\pi}{2}\right\rbrace$

4. $x\in\left\lbrace -\dfrac{\pi}{2}\;,-\dfrac{\pi}{4}\;,\dfrac{\pi}{2}\;,\dfrac{3\pi}{4}\right\rbrace$

5. $x\in\left\lbrace -\dfrac{\pi}{2}\;,0\;,\dfrac{\pi}{2}\right\rbrace$

6. $x\in\left\lbrace -\dfrac{23\pi}{36}\;,-\dfrac{11\pi}{36}\;,\dfrac{\pi}{36}\;,\dfrac{13\pi}{36}\;,\dfrac{25\pi}{36}\right\rbrace$

7. $x\in\left\lbrace -\dfrac{17\pi}{24}\;,-\dfrac{25\pi}{48}\;,-\dfrac{\pi}{48}\;,\dfrac{7\pi}{24}\;,\dfrac{23\pi}{48}\;,\dfrac{47\pi}{48}\right\rbrace$

8. $x\in \left\lbrace -\dfrac{37\pi}{60}\;,-\dfrac{13\pi}{60}\;,\dfrac{11\pi}{60}\;,\dfrac{5\pi}{12}\;,\dfrac{7\pi}{12}\;,\dfrac{59\pi}{60}\right\rbrace$

9. $x\in\left\lbrace -\dfrac{19\pi}{24}\;,-\dfrac{3\pi}{4}\;,-\dfrac{7\pi}{24}\;,\dfrac{5\pi}{24}\;,\dfrac{\pi}{4}\;,\dfrac{17\pi}{24}\right\rbrace$

10. $x\in \left\lbrace -\dfrac{59\pi}{72}\;,-\dfrac{35\pi}{72}\;,-\dfrac{17\pi}{24}\;,-\dfrac{11\pi}{72}\;,\dfrac{13\pi}{72}\;,\dfrac{37\pi}{72}\;,\dfrac{61\pi}{72}\right\rbrace$

11. $x\in\left\lbrace -\dfrac{\pi}{2}\;,\dfrac{\pi}{2}\right\rbrace$

12. $x\in \left\lbrace -\dfrac{5\pi}{12}\;,\dfrac{7\pi}{12}\right\rbrace$

Corrigé 2

Dans $]-\pi\ ;\ \pi]$

1. $x\in\left\lbrace -\dfrac{2\pi}{3}\;,-\dfrac{\pi}{3}\;,\dfrac{\pi}{3}\;,\dfrac{2\pi}{3}\right\rbrace$

2. $x\in\left\lbrace -\dfrac{\pi}{2}\;,-\dfrac{\pi}{6}\;,\dfrac{\pi}{2}\;,\dfrac{5\pi}{6}\right\rbrace$

3. $x=\left\lbrace -\dfrac{\pi}{24}+\dfrac{k\pi}{4}\;,k=-3\;,\ldots\;,4\right\rbrace$

4. $x\in\left\lbrace-\dfrac{2\pi}{3}\;,-\dfrac{\pi}{6}\;,\dfrac{\pi}{3}\;,\dfrac{5\pi}{6}\right\rbrace$

5. $x=\dfrac{\pi}{36}+\dfrac{k\pi}{6}\;,k=-6\;,\ldots\;,5$

6. $x\in\left\lbrace -\dfrac{17\pi}{24}\;,-\dfrac{13\pi}{24}\;,-\dfrac{5\pi}{24}\;,\dfrac{7\pi}{24}\;,\dfrac{11\pi}{24}\;,\dfrac{19\pi}{24}\;,\dfrac{23\pi}{24}\right\rbrace$

7. $x\in\left\lbrace -\dfrac{3\pi}{4}\;,-\dfrac{5\pi}{12}\;,-\dfrac{\pi}{12}\;,\dfrac{\pi}{4}\;,\dfrac{7\pi}{12}\;,\dfrac{11\pi}{12}\right\rbrace$

8. $x\in\left\lbrace -\dfrac{\pi}{3}\;,\dfrac{2\pi}{3}\right\rbrace$

Exercice 3

Dans $[0\ ;\ 2\pi[$

1. $\left(\cos x+1\right)\left(2\cos x+\sqrt{3}\right)=0\ :\ x\in\left\lbrace\dfrac{\pi}{6}\;,\dfrac{5\pi}{6}\;,\pi\;,\dfrac{7\pi}{6}\right\rbrace$

2. $4\cos^{2} x-3\cos x=0\ :\ x\in\left(\dfrac{\pi}{2}\;,\dfrac{3\pi}{2}\;,\arccos\dfrac{3}{4}\;,2\pi-\arccos\dfrac{3}{4}\right\rbrace$

3. $\cos 2x\left(2\cos 2x+1\right)=0\ :\ x\in\left\lbrace \dfrac{\pi}{4}\;,\dfrac{\pi}{3}\;,\dfrac{2\pi}{3}\;,\dfrac{3\pi}{4}\;,\dfrac{5\pi}{4}\;,\dfrac{4\pi}{3}\;,\dfrac{5\pi}{3}\;,\dfrac{7\pi}{4}\right\rbrace$

4. $2\cos x\left(1+\cos 4x\right)=0\ :\ x\in\left\lbrace \dfrac{\pi}{4}\;,\dfrac{\pi}{2}\;,\dfrac{3\pi}{4}\;,\dfrac{5\pi}{4}\;,\dfrac{3\pi}{2}\;,\dfrac{7\pi}{4}\right\rbrace$

5. $\cos 3x\left(2\cos 2x-1\right)=0\ :\ x\in\left\lbrace \dfrac{\pi}{6}\;,\dfrac{\pi}{2}\;,\dfrac{5\pi}{6}\;,\dfrac{7\pi}{6}\;,\dfrac{3\pi}{2}\;,\dfrac{11\pi}{6}\right\rbrace$

6. Discriminant =$-32+8\sqrt{3}< 0$ : aucune solution réelle.

7. $\sin 2x\left(2\cos x+1\right)=0\ :\ x\in\left\lbrace 0\;,\dfrac{\pi}{2}\;,\dfrac{2\pi}{3}\;,\pi\;,\dfrac{4\pi}{3}\;,\dfrac{3\pi}{2}\right\rbrace$

8. $2\sin 3x\cos x=0\ :\ x\in\left\lbrace 0\;,\dfrac{\pi}{3}\;,\dfrac{\pi}{2}\;,\dfrac{2\pi}{3}\;,\pi\;,\dfrac{4\pi}{3}\;,\dfrac{3\pi}{2}\;,\dfrac{5\pi}{3}\right\rbrace$

9. $1+\dfrac{2}{\sqrt{3}}\tan x-\tan^{2} x=0\Longleftrightarrow \tan x=\sqrt{3}\text{ ou }-\dfrac{1}{\sqrt{3}}\ :\ x\in\left\lbrace \dfrac{\pi}{3}\;,\dfrac{5\pi}{6}\;,\dfrac{4\pi}{3}\;,\dfrac{11\pi}{6}\right\rbrace$

10. $\left(\tan x+1\right)\left(\sqrt{3}\tan x+1\right)=0\ :\ x\in\left\lbrace \dfrac{3\pi}{4}\;,\dfrac{5\pi}{6}\;,\dfrac{7\pi}{4}\;,\dfrac{11\pi}{6}\right\rbrace$

11. $\left(\tan x-1\right)\left(\tan x+\sqrt{3}\right)=0\ :\ x\in\left\lbrace \dfrac{\pi}{4}\;,\dfrac{2\pi}{3}\;,\dfrac{5\pi}{4}\;,\dfrac{5\pi}{3}\right\rbrace$

(La question $5^{\circ}$ apparaît en double dans l'énoncé imprimé ; seule la première version, ci-dessus, a été corrigée.)

Corrigé 4

1. $\sin x=\dfrac{4}{5}\;,\tan x=-\dfrac{4}{3}$

2. $\sin x=-\dfrac{2}{3}\;,\tan x=-\dfrac{2\sqrt{5}}{5}$

3. $\sin x=\dfrac{5}{13}\;,\tan x=-\dfrac{5}{12}$

Corrigé 5

1. $\sin x=\dfrac{2\sqrt{5}}{5}\;,\cos x=-\dfrac{\sqrt{5}}{5}$

2. $\sin x=-\dfrac{11}{61}\;,\cos x=-\dfrac{11}{61}\;,\cos x=-\dfrac{60}{61}$

3. $\sin x=-\dfrac{8}{17}\;,\cos x=\dfrac{15}{17}$

Corrigé 6

1. $\cos x=-\dfrac{\sqrt{3}}{3}\;,\tan x=-\sqrt{2}$

2. $\cos x=-\dfrac{3}{5}\;,\tan x=\dfrac{4}{3}$

3. $\cos x=\dfrac{\sqrt{7}}{3}\;,\tan x=-\sqrt{\dfrac{2}{7}}$

Corrigé 7 

$\tan 2x=\dfrac{2\left(\sqrt{3}-2\right)}{1-\left(\sqrt{3}-2\right)^{2}}=-\dfrac{\sqrt{3}}{3}$

comme $2x\in]-\pi\;,0[\;,2x=-\dfrac{\pi}{6}$ donc $x=-\dfrac{\pi}{12}$

Corrigé 8

$\sin x=\dfrac{ \sqrt{2}+\sqrt{6}}{4}=\sin\left(\dfrac{5\pi}{12}\right)$, avec $x\in\left] 0\;,\dfrac{\pi}{2}\right[$

Donc $x=\dfrac{5\pi}{12}$ et $\cos 2x=-\dfrac{\sqrt{3}}{2}$

Corrigé 9

Solutions données dans $[0\;,2\pi[$ puis dans $]-\pi\;,\pi]$

1. $\left[0\ ;\ \dfrac{2\pi}{3}\right[\cup\left]\dfrac{4\pi}{3}\;,2\pi\right[\ ;\ \left]-\dfrac{2\pi}{3}\;,\dfrac{2\pi}{3}\right[$ 

2. $\left[\dfrac{\pi}{4}\;,\dfrac{7\pi}{4}\right]\ ;\ \left]-\pi\;,-\dfrac{\pi}{4}\right]\cup \left[\dfrac{\pi}{4}\;,\pi \right]$

3. $\left[\dfrac{\pi}{6}\;,\dfrac{5\pi}{6}\right]$ ; même résultat

4. $\left[0\;,\dfrac{\pi}{4}\right[ \cup\left]\dfrac{2\pi}{3}\;,2\pi \right[\ ;\ \left]-\pi\;,\dfrac{\pi}{3}\right[\cup \left]\dfrac{2\pi}{3}\;,\pi \right]$

5. $\left]\dfrac{\pi}{4}\;,\dfrac{\pi}{2}\right[ \cup\left]\dfrac{5\pi}{4}\;,\dfrac{3\pi}{2}\right [\ ;\ \left]-\dfrac{3\pi}{4}\;,-\dfrac{\pi}{2}\right[\cup\left]\dfrac{\pi}{4}\;,\dfrac{\pi}{2}\right[$

6. $\left[0\;,\dfrac{\pi}{3}\right] \cup\left]\dfrac{\pi}{2}\;,\dfrac{4\pi}{3}\right]\cup\left]\dfrac{3\pi}{2}\;,2\pi\right[\ ;\ \left]-\pi\;,-\dfrac{5\pi}{6}\;,\right]\cup\left]-\dfrac{\pi}{2}\;,\dfrac{\pi}{3}\right]\cup\left]\dfrac{\pi}{2}\;,\pi \right]$

7. Factorisation : $4\left(\cos x+\dfrac{1}{2}\right)\left(\cos x-\dfrac{\sqrt{2}}{2}\right)> 0$

Solutions : $\left[0\;,\dfrac{\pi}{4}\right[\cup\left]\dfrac{2\pi}{3}\;,\dfrac{4\pi}{3}\right]\cup\left]\dfrac{7\pi}{4}\;,2\pi\right[\ ;\ \left]-\pi\;,-\dfrac{2\pi}{3}\right[\cup\left]-\dfrac{\pi}{4}\;,\dfrac{\pi}{4}\right[\cup\left]-\dfrac{2\pi}{3}\;,\pi\right]$

8.Factorisation : $4\left(\sin x-\dfrac{1}{2}\right)\left(\sin x-\dfrac{\sqrt{3}}{2}\right) > 0: $ Solution : $\left[\dfrac{\pi}{6}\;,\dfrac{\pi}{3}\right]\cup\left[\dfrac{2\pi}{3}\;,\dfrac{5\pi}{6}\right]$ (dans les deux intervalles)

9. Factorisation : $4\left(\sin x-\dfrac{1}{2}\right)\left(\sin x+\dfrac{\sqrt{2}}{2}\right)\leq 0$

solution : $\left[\dfrac{7\pi}{6}\;,\dfrac{5\pi}{4}\right]\cup\left[\dfrac{7\pi}{4}\;,\dfrac{11\pi}{6}\right]\ ;\ \left[-\dfrac{5\pi}{6}\;,-\dfrac{3\pi}{4}\right]\cup\left[-\dfrac{\pi}{4}\;,-\dfrac{\pi}{6}\right]$

Corrigé 10

a. $\alpha-5\pi\ :\ \left(\sin\;,\cos\;,\tan\right)=\left(-\sin \alpha\;,-\cos\alpha\;,\tan\alpha\right)$

b. $-\alpha-\pi\ :\ \left(\sin\alpha\;,\cos\alpha\;,-\tan\alpha\right)$

c. $-\alpha-2\pi\ :\ \left(-\sin \alpha\;,\cos\alpha\;,-\tan\alpha \right)$

d. $-\alpha-\dfrac{\pi}{2}\ :\ \left(-\cos\alpha\;,-\sin\alpha\;,cotan  \alpha\right)$

e. $\alpha-\dfrac{9\pi}{2}\ :\ \left(-\cos\alpha\;,-\sin\alpha\;,-cotan  \alpha\right)$

f. $-\alpha+\dfrac{5\pi}{2}\ :\ \left(\cos\alpha\;,\sin\alpha\;,cotan  \alpha\right)$

Corrigé 11

a. $\pi\ :\ \left(0\;,-1\;,0\right)\ ;\ -5\pi\ :\ (0\;,-1\;,0)\ ;\ \dfrac{3\pi}{2}\ ;\ \left(-1\;,0\;,\text{non définie}\right)$

b. $\dfrac{3\pi}{4}\ :\ \left(\dfrac{\sqrt{2}}{2}\;,-\dfrac{\sqrt{2}}{2}\;,-1\right)\ ;\ -\dfrac{7\pi}{4}\ :\ \left(\dfrac{\sqrt{2}}{2}\;,\dfrac{\sqrt{2}}{2}\;,1\right)\ ;\ \dfrac{15\pi}{4}\ :\ \left(-\dfrac{\sqrt{2}}{2}\;,\dfrac{2}{2}\;,-1\right)$

c. $-\dfrac{2\pi}{3}\ ;\ \left(-\dfrac{\sqrt{3}}{2}\;,-\dfrac{1}{2}\;,\sqrt{3}\right)\ ;\ \dfrac{5\pi}{3}\ ;\ \left(-\dfrac{\sqrt{3}}{2}\;,\dfrac{1}{2}\;,-\sqrt{3}\right)\ ;\ \dfrac{8\pi}{3}\ ;\ \left(\dfrac{\sqrt{3}}{2}\;,-\dfrac{1}{2}\;,-\sqrt{3}\right)$

d. $\dfrac{7\pi}{6}\ ;\ \left(-\dfrac{1}{2}\;,-\dfrac{\sqrt{3}}{2}\;,\dfrac{\sqrt{3}}{3}\right)\ ;\ \dfrac{13\pi}{6}\ ;\ \left(\dfrac{1}{2}\;,\dfrac{\sqrt{3}}{2}\;,\dfrac{\sqrt{3}}{3}\right)\ ; \dfrac{31\pi}{6}\ ;\ \left(-\dfrac{1}{2}\;,-\dfrac{\sqrt{3}}{2}\;,\dfrac{\sqrt{3}}{3}\right)$

Corrigé 12

$A=0$

$B=cotan \alpha$

$C=-\dfrac{1}{3}\tan \left(3\alpha\right)$ si $k$ est pair ; $C=-\dfrac{1}{5}\tan\left(3\alpha\right)$ si $k$ est impaire

Corrigé 13

1. $f(x)=1$

2. $f(x)=\dfrac{3}{2}$

3. Anomalie de l'énoncé : tel qu'imprime, la fonction est en réalité $f(x)=\sin^{2} x+\sin^{2}\left(\dfrac{\pi}{3}+x\right)+\sin^{2}\left(\dfrac{2\pi}{3}-x\right)$, ce qui donne $f(x)=\dfrac{3}{2}+\dfrac{\sqrt{3}}{2}\sin 2x$ : cette fonction n'est pas constante $\left(\text{par exemple }f(0)=\dfrac{3}{2}\neq
 f\left(\dfrac{\pi}{4}\right)=\dfrac{3+\sqrt{3}}{2}\right)$ Si le dernier terme était $\sin^{2}\left(\dfrac{2\pi}{3}+x\right)$, on aurait bien $f(x)=\dfrac{3}{2}$
 
Corrigé 14

1. $\sqrt{2}\sin\left(\dfrac{\pi}{4}+x\right)=\cos x+\sin x$

2. $\sqrt{2}\sin\left(\dfrac{\pi}{4}-x\right)=\cos x-\sin x=\sqrt{2}\cos\left(\dfrac{\pi}{4}+x\right)$

3. Par les formules somme-produit : $\cos x+\cos 2x+\cos 3x+\cos 4x=4\cos\dfrac{x}{2}\cos x\cos\dfrac{5x}{2}$

4. $\sin x+\sin 2x+\sin 3x+\sin 4x=4\cos\dfrac{x}{2}\cos x\sin\dfrac{5x}{2}$

5. $4\sin x\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{3}+x\right)=3x\sin x-4\sin^{3}x=\sin 3x$