4.2.1. Қарапайым тригонометриялық теңдеулер

$${0,5x = - \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$ $${x = - \pi + 4\pi n;\quad n \in Z}$$

$${0,5x = - \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$ $${x = - \pi + 4\pi n;\quad n \in Z}$$

Жауабы: ${\left\{ { - \pi + 4\pi n} \right\}}$

$${x = \pm \arccos \frac{1}{2} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${40^\circ \le 60^\circ + 360^\circ n \le 1050^\circ }$$ $${640^\circ \le 360^\circ n \le 910^\circ }$$ $${1,7 \le n \le 2,52}$$ $${n = 2}$$

$${700^\circ \le - 60^\circ + 360^\circ n \le 1050^\circ }$$ $${760^\circ \le 360^\circ n \le 1110^\circ }$$ $${21 \le n \le 3,08}$$ $${n = 3.}$$ $${n = 2,}$$ $${\quad x = 60^\circ + 360^\circ \cdot 2 = 780^\circ }$$ $${n = 3,}$$ $${\quad x = - 60^\circ + 360^\circ \cdot 3 = 1020^\circ }$$

Жауабы: Ең үлкен шешімі: ${x = 1020^\circ }$

$${ - \sin x - \sin x = - 1}$$ $${ - 2\sin x = - 1}$$

$${\sin x = \frac{1}{2}}$$ $${x = {{( - 1)}^k}\arcsin \frac{1}{2} + \pi k;\quad k \in Z}$$

Жауабы: ${x = {{( - 1)}^k}\frac{\pi }{6} + \pi k;\quad k \in Z}$

$${ - 2\sqrt 3 \tg x = - 6}$$ $${\tg x = \frac{{ - 6}}{{ - 2\sqrt 3 }}}$$

$${\tg x = \sqrt 3 }$$ $${x = \arctg \sqrt 3 + \pi n;\quad n \in Z}$$

Жауабы: ${x = \frac{\pi }{3} + \pi n;\quad n \in Z}$

$${x - \frac{\pi }{4} = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi n;}$$ $${\quad n \in Z}$$

$${x - \frac{\pi }{4} = \pm \frac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{{2\pi }}{3} + \frac{\pi }{4} + 2\pi n;\quad n \in Z}$$

Жауабы: ${ \pm \frac{{2\pi }}{3} + \frac{\pi }{4} + 2\pi n;\quad n \in Z}$

$${ - \tg \left( {\frac{x}{2} - \frac{\pi }{4}} \right) = - 1}$$ $${\tg \left( {\frac{x}{2} - \frac{\pi }{4}} \right) = 1}$$ $${\frac{x}{2} - \frac{\pi }{4} = \arctg 1 + \pi n;\quad n \in Z}$$

$${\frac{x}{2} - \frac{\pi }{4} = \frac{\pi }{4} + \pi n;\quad n \in Z}$$ $${\frac{x}{2} = \frac{\pi }{4} + \frac{\pi }{4} + \pi n;\quad n \in Z}$$ $${\frac{x}{2} = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $$x = \pi + 2\pi n;\quad n \in Z$$

Жауабы: $x = \pi \left( {1 + 2n} \right);\quad n \in Z$

$${\frac{{4x}}{3} + \frac{\pi }{3} = 2\pi n;\quad n \in Z}$$ $${\frac{{4x}}{3} = - \frac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${4x = - \pi + 6\pi n;\quad n \in Z}$$

Жауабы: ${x = - \frac{\pi }{4} + \frac{{3\pi n}}{2};\quad n \in Z}$

Анықталу облысы: $$\cos x \gt 0$$ $${\cos x = 10}$$ $${\cos x \in [ - 1;1]}$$

Жауабы: $\emptyset$

$$ - \sin \left( {x - \frac{\pi }{4}} \right) = \frac{1}{2}$$ $$\sin \left( {x - \frac{\pi }{4}} \right) = - \frac{1}{2}$$

$${x - \frac{\pi }{4} = {{( - 1)}^{n + 1}}\arcsin \frac{1}{2} + \pi n;}$$ $${\quad n \in Z}$$ $${x = {{( - 1)}^{n + 1}}\frac{\pi }{6} + \frac{\pi }{4} + \pi n;}$$ $${\quad n \in Z}$$

Жауабы: ${{{( - 1)}^{n + 1}}\frac{\pi }{6} + \frac{\pi }{4} + \pi n;\quad n \in Z}$

$$2\sin x\cos x = 1$$ $$\sin 2x = 1$$ $${2x = \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$

Жауабы: ${x = \frac{\pi }{4} + \pi n;\quad n \in Z}$

$${\frac{{\sqrt 3 }}{2}\tg \left( {2x - \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}}$$ $${\tg \left( {2x - \frac{\pi }{4}} \right) = 1}$$ $${2x - \frac{\pi }{4} = \arctg 1 + \pi n;\quad n \in Z}$$

$${2x - \frac{\pi }{4} = \frac{\pi }{4} + \pi n;\quad n \in Z}$$ $${2x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$

Жауабы: ${x = \frac{\pi }{4} + \frac{{\pi n}}{2};\quad n \in Z}$

$${3x - 10^\circ = \arctg 0 + \pi n;\quad n \in Z}$$ $${3x = 10^\circ + \pi n;\quad n \in Z}$$

$${x = \frac{{10^\circ }}{3} + \frac{{90^\circ n}}{3};\quad n \in Z}$$ $${x = \frac{{9^\circ 60′}}{3} + 30^\circ n;\quad n \in Z}$$

Жауабы: ${x = 3^\circ 20′ + 30^\circ n;\quad n \in Z}$

Анықталу облысы: $$\cos x \gt 0$$ $$\cos x = {3^0}$$ $$\cos x = 1$$

Жауабы: $x = 2\pi n;\quad n \in Z$

$${\cos \frac{x}{2} + 1 = 0}$$ $${\cos \frac{x}{2} = - 1}$$ $${\frac{x}{2} = \pi + 2\pi n;\quad n \in Z}$$ $${x = 2\pi + 4\pi n;\quad n \in Z}$$

$${{{\sin }^2}\frac{x}{2} + 2 = 0}$$ $${{{\sin }^2}\frac{x}{2} = - 2}$$ $$\emptyset $$

Жауабы: $2\pi + 4\pi n;\quad n \in Z$

$${x - 1 = \pm \arccos \frac{{\sqrt 3 }}{2} + 2\pi n;}$$ $${\quad n \in Z}$$

$${x - 1 = \pm \frac{\pi }{6} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{6} + 1 + 2\pi n;\quad n \in Z}$$

Жауабы: ${1 \pm \frac{\pi }{6} + 2\pi n;\quad n \in Z}$

$${\frac{{2\cos x}}{{\cos x}} + \frac{{3\sin x}}{{\cos x}} = \frac{0}{{\cos x}}}$$ $${2 + 3\tg x = 0}$$ $${3\tg x = - 2}$$

$${\tg x = - \frac{2}{3}}$$ $${x = \arctg \left( { - \frac{2}{3}} \right) + \pi n;\quad n \in Z}$$ $${x = - \arctg \frac{2}{3} + \pi n;\quad n \in Z}$$

Жауабы: ${ -\arctg \frac{2}{3} + \pi n;\quad n \in Z}$

$${{3^{|\sin x - 1|}} = {3^2}}$$ $${|\sin x - 1| = 2}$$ $${\sin x - 1 = 2}$$ $${\sin x = 3}$$ $${\sin x \in [ - 1;1]}$$ $$\emptyset$$

$${\sin x - 1 = - 2}$$ $${\sin x = - 1}$$ $${x = - \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$

Жауабы: ${ - \frac{\pi }{2} + 2\pi n;\quad n \in Z}$

$${{2^{2\sin x\cos x}} = {2^{\frac{1}{2}}}}$$ $${2\sin x\cos x = \frac{1}{2}}$$ $${\sin 2x = \frac{1}{2}}$$ $${2x = {{( - 1)}^n}\arcsin \frac{1}{2} + \pi n;\quad n \in Z}$$

$${2x = {{( - 1)}^n}\frac{\pi }{6} + \pi n,}$$ $${\quad n \in Z}$$ $${x = {{( - 1)}^n}\frac{\pi }{{12}} + \frac{{\pi n}}{2};\quad n \in Z}$$

Жауабы: ${{{( - 1)}^n}\frac{\pi }{{12}} + \frac{{\pi n}}{2};\quad n \in Z}$

$${2\sin \frac{{\pi x}}{3} = - 1}$$ $${\sin \frac{{\pi x}}{3} = - \frac{1}{2}}$$ $${\frac{{\pi x}}{3} = {{( - 1)}^{n + 1}}\arcsin \frac{1}{2} + \pi n;}$$ $${\quad n \in Z}$$ $${\frac{{\pi x}}{3} = {{( - 1)}^{n + 1}}\frac{\pi }{6} + \pi n;\quad n \in Z}$$ $${\frac{x}{3} = {{( - 1)}^{n + 1}}\frac{1}{6} + n;\quad n \in Z}$$ $${x = {{( - 1)}^{n + 1}}\frac{1}{2} + 3n;\quad n \in Z}$$ $${2 \lt {{( - 1)}^{n + 1}}\frac{1}{2} + 3n \lt 4}$$

$${2 \lt - \frac{1}{2} + 3n \lt 4}$$ $${2,5 \lt 3n \lt 4,5}$$ $${0,83 \lt n \lt 1,5}$$ $${n = 1.}$$ $${2 \lt \frac{1}{2} + 3n \lt 4}$$ $${1,5 \lt 3n \lt 3,5}$$ $${0,5 \lt n \lt 1,16}$$ $${n = 1.}$$ $$x = {( - 1)^{1 + 1}}\frac{1}{2} + 3 \cdot 1 =$$ $$= \frac{1}{2} + 3 = 3,5$$

Жауабы: $\quad x = 3,5$

$${5^1} \cdot {5^{{{\log }_5}\cos x}} = 2,5$$ Анықталу облысы: ${\cos x \gt 0}$ $${5 \cdot \cos x = 2,5}$$

$${\cos x = \frac{1}{2}}$$ $${x = \pm \arccos \frac{1}{2} + 2\pi n;\quad n \in Z}$$

Жауабы: ${x = \pm \frac{\pi }{3} + 2\pi n;\quad n \in Z}$

$${\pi \cos 3x = \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$ $${\cos 3x = \frac{1}{2} + 2n;\quad n \in Z}$$ $${\cos 3x \in [ - 1;1]}$$ $${ - 1 \le \frac{1}{2} + 2n \le 1}$$ $${ - 1,5 \le 2n \le 0,5}$$ $${ - 0,75 \le n \le 0,25,\quad n = 0}$$

$${\cos 3x = \frac{1}{2} + 2 \cdot 0}$$ $${\cos 3x = \frac{1}{2}}$$ $${3x = \pm \arccos \frac{1}{2} + 2\pi n;\quad n \in Z}$$ $${3x = \pm \frac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{9} + \frac{{2\pi n}}{3};\quad n \in Z}$$

Жауабы: ${ \pm \frac{\pi }{9} + \frac{{2\pi n}}{3};\quad n \in Z}$

$${{2^{3{{\sin }^2}x}} = {2^{{{\cos }^2}x}}}$$ $${3{{\sin }^2}x = {{\cos }^2}x}$$ $${3{{\sin }^2}x - \left( {1 - {{\sin }^2}x} \right) = 0}$$ $${4{{\sin }^2}x - 1 = 0}$$ $${4 \cdot \frac{{1 - \cos 2x}}{2} - 1 = 0}$$ $${2(1 - \cos 2x) - 1 = 0}$$ $${2 - 2\cos 2x - 1 = 0}$$

$${2\cos 2x = 1}$$ $${\cos 2x = \frac{1}{2}}$$ $${2x = \pm \arccos \frac{1}{2} + 2\pi n;\quad n \in Z}$$ $${2x = \pm \frac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{6} + \pi n;\quad n \in Z}$$

Жауабы: ${ \pm \frac{\pi }{6} + \pi n;\quad n \in Z}$

$${3\cos \pi x - \pi = 0}$$ $${3\cos \pi x = \pi }$$ $${\cos \pi x = \frac{\pi }{3}}$$ $${\cos \pi x \in [ - 1;1]}$$ $$\emptyset $$ $${2\sin \pi x - \sqrt 3 = 0}$$ $${\sin \pi x = \frac{{\sqrt 3 }}{2}}$$ $${\pi x = {{( - 1)}^n}\arcsin \frac{{\sqrt 3 }}{2} + \pi n;}$$ $${\quad n \in Z}$$ $${\pi x = {{( - 1)}^n}\frac{\pi }{3} + \pi n;\quad n \in Z}$$

$$x = {( - 1)^n}\frac{1}{3} + n;\quad n \in Z$$ $$n - \text{тақ болса},\quad - \frac{1}{3} + n \gt 0$$ $${n \gt \frac{1}{3},\quad n = 1}$$ $${n = 1,\quad x = - \frac{1}{3} + 1 = \frac{2}{3}}$$ $$n - \text{жұп болса} ,\quad \frac{1}{3} + n \gt 0$$ $$n \gt - \frac{1}{3},\quad n = 0$$ $${n = 0,\quad x = \frac{1}{3} + 0 = \frac{1}{3}}$$

Жауабы: Ең кіші оң түбірі ${\frac{1}{3}}$

$${\cos x = 0}$$ $${x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${\sin x - 1 \ne 0.}$$ $${\sin x \ne 1}$$ $${x \ne \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$

Жауабы: $x = - \frac{\pi }{2} + 2\pi n;\quad n \in Z$

$${2\cos \frac{{\pi x}}{{15}} = - 1}$$ $${\cos \frac{{\pi x}}{{15}} = - \frac{1}{2}}$$ $${\frac{{\pi x}}{{15}} = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi n;}$$ $${\quad n \in Z}$$

$${\frac{{\pi x}}{{15}} = \pm \frac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ $${\frac{x}{{15}} = \pm \frac{2}{3} + 2n;\quad n \in Z}$$ $${x = \pm 10 + 30n;\quad n \in Z}$$

Жауабы: ${x = \pm 10 + 30n;\quad n \in Z}$

$${\cos x - \frac{1}{2} = 0}$$ $${\cos x = \frac{1}{2}}$$ $${x = \pm \frac{\pi }{3} + 2\pi n,\quad n \in Z}$$ $${\sin x - \frac{{\sqrt 3 }}{2} \ne 0}$$ $${\sin x \ne \frac{{\sqrt 3 }}{2}}$$ $${x \ne {{( - 1)}^n}\arcsin \frac{{\sqrt 3 }}{2} + \pi n;}$$ $${\quad n \in Z}$$

$$x \ne {( - 1)^n}\frac{\pi }{3} + \pi n;\quad n \in Z$$

Жауабы: $x = - \frac{\pi }{3} + 2\pi n;\quad n \in Z$

$${\sin 2x = 1}$$ $${2x = \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$ $${x = \frac{\pi }{4} + \pi n;\quad n \in Z}$$ $$x = \frac{\pi }{4} + \frac{{\pi n}}{2},\quad n \in Z$$ $${\sin 2x = - 1}$$ $${2x = - \frac{\pi }{2} + 2\pi n;\quad n \in Z}$$

$${x = - \frac{\pi }{4} + \pi n;\quad n \in Z}$$

Жауабы: $x = \frac{\pi }{4} + \frac{{\pi n}}{2},\quad n \in Z$

$$\tg 3x = 0$$ $$3x = \arctg 0 + \pi n;\quad n \in Z$$ $$x = \frac{{\pi n}}{3}\quad n \in Z$$ $$\sin x \ne 0$$ $$x \ne \pi n;\quad n \in Z$$

Жауабы: $x = \pm \frac{\pi }{3} + \pi n;\quad n \in Z$

$${\frac{{1 + \cos 4x}}{2} = \frac{1}{2}}$$ $${1 + \cos 4x = 1}$$ $${\cos 4x = 0}$$

$${4x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${x = \frac{\pi }{8} + \frac{{\pi n}}{4};\quad n \in Z}$$

Жауабы: ${\frac{\pi }{8} + \frac{{\pi n}}{4};\quad n \in Z}$

$${\frac{{1 - \cos 4x}}{2} = \frac{1}{2}}$$ $${1 - \cos 4x = 1}$$ $${\cos 4x = 0}$$

$${4x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${x = \frac{\pi }{8} + \frac{{\pi n}}{4};\quad n \in Z}$$

Жауабы: ${\frac{\pi }{8} + \frac{{\pi n}}{4};\quad n \in Z}$