Тригонометриялық теңдеулер (Рустюмова 4.2.2 A (1-21))

Шешуі: $$2\sin \frac{{3x + x}}{2}\cos \frac{{3x - x}}{2} = 0$$ $$\sin 2x \cdot \cos x = 0$$ $$1) \quad \sin 2x=0 $$ $$2x=\pi k; \quad k \in Z$$ $$x \in Z $$ $$ x= \frac{\pi k}{2}; \quad k \in Z $$ $$2) \quad \cos x =0 $$ $$x=\frac{\pi}{2}+\pi k; $$ Жауабы: $x=\frac{\pi k}{2}; \quad k \in Z$

Шешуі: $$\sqrt 3 \sin x\cos x - {\sin ^2}x = 0$$ $$\sin x(\sqrt 3 \cos x - \sin x) = 0$$ $$1)\quad \sin x = 0$$ $$x=\pi n,\quad n \in Z;$$ $$2)\quad \sqrt 3 \cos x - \sin x = 0,\quad \left| { \div \,\cos x} \right.$$ $$\frac{{\sqrt 3 \cos x}}{{\cos x}} - \frac{{\sin x}}{{\cos x}} = \frac{0}{{\cos x}}$$ $$\sqrt 3 - \tg x = 0$$ $$\tg x = \sqrt 3 $$ $${x = \arctg \sqrt 3 + \pi k;\quad k \in Z}$$ $${x = \frac{\pi }{3} + \pi k;\quad k \in Z}$$ Жауабы:$\quad {\pi n;\quad \frac{\pi }{3} + \pi k;\quad n,k \in Z}$

Шешуі: $${{{\sin }^2}x - \frac{1}{2} \cdot 2\sin x\cos x = 0}$$ $${{{\sin }^2}x - \sin x\cos x = 0}$$ $${\sin x(\sin x - \cos x) = 0}$$ $$1)\quad \sin x = 0,\quad x = \pi k;\quad k \in Z$$ $$2)\quad \sin x - \cos x = 0,\quad \left| { \div \cos x} \right.$$ $${\frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\cos x}} = 0}$$ $${\tg x - 1 = 0}$$ $${\tg x = 1}$$ $$ x = \frac{\pi }{4} + \pi n;\quad n \in Z$$ Жауабы: $\quad \pi k;\quad \frac{\pi }{4} + \pi n;\quad n,k \in Z$

Шешуі: $$\cos x(3\sin x - 2\cos x) = 0$$ $$1) \cos x = 0,\quad x = \frac{\pi }{2} + \pi k, k \in Z$$ $$2)\quad 3\sin x - 2\cos x = 0$$ $${\frac{{3\sin x}}{{\cos x}} - \frac{{2\cos x}}{{\cos x}} = 0}$$ $${3\tg x - 2 = 0}$$ $${3\tg x = 2}$$ $${\tg x = \frac{2}{3}}$$ $$x = \arctg \frac{2}{3} + \pi n;\quad n \in Z$$ Жауабы: $ x = \frac{\pi }{2} + \pi k; \arctg \frac{2}{3} + \pi n; n,k \in Z$

Шешуі: $${\frac{1}{2}(\cos 6x + \cos 4x) = \cos 4x}$$ $${\cos 6x + \cos 4x = 2\cos 4x}$$ $${\cos 6x - \cos 4x = 0}$$ $${ - 2\sin \frac{{6x + 4x}}{2}\sin \frac{{6x - 4x}}{2} = 0}$$ $${\sin 5x\sin x = 0}$$ $$1)\quad \sin 5x = 0;\quad 5x = \pi k;\quad x = \frac{{\pi k}}{5},\quad k \in Z$$ $$2)\quad \sin x = 0;\quad x = \pi n;\quad n \in Z$$ Жауабы: $\quad \frac{{\pi k}}{5},\quad k \in Z$

Шешуі: $${\sin x\sin 2x + \cos (2x + x) = 0}$$ $${\sin x\sin 2x + \cos 2x\cos x - \sin 2x\sin x = 0}$$ $${\cos 2x\cos x = 0}$$ $${1)\quad \cos 2x = 0}$$ $${2x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${x = \frac{\pi }{4} + \frac{{\pi n}}{2};\quad n \in Z}$$ $$2)\quad \cos x = 0,\quad x \in \frac{\pi }{2} + \pi k;\quad k \in Z$$ Жауабы: $\quad x = \frac{\pi }{4} + \frac{{\pi n}}{2};\quad x \in \frac{\pi }{2} + \pi k;\quad n,k \in Z$

Шешуі: $${\cos x\cos 4x = \cos (4x + x)}$$ $${\cos x\cos 4x = \cos 4x\cos x - \sin 4x\sin x}$$ $${\cos x\cos 4x - \cos 4x\cos x + \sin 4x\sin x = 0}$$ $${\sin 4x\sin x = 0}$$ $$1)\quad \sin x = 0$$ $$4x = \pi n;\quad n \in Z$$ $$x \in \frac{{\pi n}}{4};\quad n \in Z$$ $$2)\quad \sin x = 0$$ $$x = \pi k;\quad k \in Z$$ Жауабы: $\quad \frac{{\pi n}}{4};\quad n \in Z$

Шешуі: $${\cos x(3\sin x - 5\cos x) = 0}$$ $$1)\quad \cos x = 0$$ $$x = \frac{\pi }{2} + \pi n;\quad n \in Z$$ $$2)\quad 3\sin x - 5\cos x = 0$$ $${\frac{{3\sin x}}{{\cos x}} - \frac{{5\cos x}}{{\cos x}} = 0}$$ $${3\tg x - 5 = 0}$$ $${\tg x = \frac{5}{3}}$$ $$x = \arctg \frac{5}{3} + \pi ;\quad k \in Z$$ Жауабы:$\quad \frac{\pi }{2} + \pi n;\quad \arctg \frac{5}{3} + \pi ;\quad n,k \in Z$

Шешуі: $${\cos (3x + x) = - \frac{1}{2}}$$ $${\cos 4x = - \frac{1}{2}}$$ $${4x = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi n;\quad n \in Z}$$ $${4x = \pm \frac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{6} + \frac{{\pi n}}{2};\quad n \in Z}$$ Жауабы:$\quad {x = \pm \frac{\pi }{6} + \frac{{\pi n}}{2};\quad n \in Z}$

Шешуі: $${(\sin x + \sin 3x) + \sin 2x = 0}$$ $${2\sin \frac{{x + 3x}}{2}\cos \frac{{3x - x}}{2} + \sin 2x = 0}$$ $${2\sin 2x\cos x + \sin 2x = 0}$$ $$\sin 2x(2\cos x + 1) = 0$$ $$1)\quad \sin 2x = 0$$ $${2x = \pi k;\quad k \in Z}$$ $${x = \frac{{\pi k}}{2};\quad k \in Z}$$ $$2)\quad 2\cos x + 1 = 0$$ $${2\cos x = - 1}$$ $${\cos x = - \frac{1}{2}}$$ $${x = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{{2\pi }}{3} + 2\pi n;\quad n \in Z}$$ Жауабы: $\quad \frac{{\pi k}}{2};\quad \pm \frac{{2\pi }}{3} + 2\pi n;\quad n,k \in Z$

Шешуі: $${2\sin x\cos x - 2\sqrt 3 {{\cos }^2}x = 0}$$ $${\cos x(2\sin x - 2\sqrt 3 \cos x) = 0}$$ $$1)\quad \cos x = 0$$ $${x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $$2)\quad 2\sin x - 2\sqrt 3 \cos x = 0$$ $${\frac{{2\sin x}}{{\cos x}} - \frac{{2\sqrt 3 \cos x}}{{\cos x}} = \frac{0}{{\cos x}}}$$ $${2\tg x - 2\sqrt 3 = 0}$$ $${2\tg x = 2\sqrt 3 }$$ $${\tg x = \sqrt 3 }$$ $${x = \frac{\pi }{3} + \pi k;\quad k \in Z}$$ Жауабы: $\quad \frac{\pi }{2} + \pi n;\quad \frac{\pi }{3} + \pi k;\quad n,k \in Z$

Шешуі: $$2\sin \frac{{x + 3x}}{2}\cos \frac{{x - 3x}}{2} - 2\sin 2x = 0$$ $$2\sin 2x\cos x - 2\sin 2x = 0$$ $$\sin 2x(2\cos x - 2) = 0$$ $${1)\quad \sin 2x = 0}$$ $${2x = \pi n;\quad n \in Z}$$ $${x = \frac{{\pi n}}{2};\quad n \in Z}$$ $${2)\quad 2\cos x - 2 = 0}$$ $${2\cos x = 2}$$ $${\cos x = 1}$$ $${x = \frac{\pi }{2} + \pi k;\quad k \in Z}$$ Жауабы: $\quad \frac{{\pi n}}{2};\quad n \in Z$

Шешуі: $${{{\sin }^2}x + \frac{1}{2} \cdot 2\sin x\cos x - 1 = 0}$$ $${\sin x\cos x - {{\cos }^2}x = 0}$$ $${\cos x(\sin x - \cos x) = 0}$$ $${1)\quad \cos x = 0}$$ $${x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${2)\quad \sin x - \cos x = 0}$$ $${\frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\cos x}} = \frac{0}{{\cos x}}}$$ $${\tg x - 1 = 0}$$ $${\tg x = 1}$$ $${x = \frac{\pi }{4} + \pi k;\quad k \in Z}$$ Жауабы:$$\quad \frac{\pi }{2} + \pi n;\quad \frac{\pi }{4} + \pi k;\quad n,k \in Z$$

Шешуі: $${\sqrt 3 {{\cos }^2}x - 0,5 \cdot 2\sin x\cos x = 0}$$ $${\sqrt 3 {{\cos }^2}x - \sin x\cos x = 0}$$ $${\cos x(\sqrt 3 \cos x - \sin x) = 0}$$ $$1)\quad \cos x = 0;\quad x = \frac{\pi }{2} + \pi n;\quad n \in Z$$ $${2)\quad \sqrt 3 \cos x - \sin x = 0}$$ $${\frac{{\sqrt 3 \cos x}}{{\cos x}} - \frac{{\sin x}}{{\cos x}} = \frac{0}{{\cos x}}}$$ $${\sqrt 3 - \tg x = 0}$$ $${\tg x = \sqrt 3 }$$ $${x = \frac{\pi }{3} + \pi k;\quad k \in Z}$$ Жауабы:$\quad \frac{\pi }{2} + \pi n;\quad \frac{\pi }{3} + \pi k;\quad n,k \in Z$

Шешуі: $${2\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right) = 1}$$ $${2\cos 2x = 1}$$ $${\cos 2x = \frac{1}{2}}$$ $${2x = \pm \arccos \frac{1}{2} + 2\pi n;\quad n \in Z}$$ $${2x = \pm \frac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{6} + \pi n;\quad n \in Z}$$ Жауабы: $${\quad x = \pm \frac{\pi }{6} + \pi n;\quad n \in Z}$$

Шешуі: $${2\left( {\cos x\cos \frac{\pi }{6} - \sin x\sin \frac{\pi }{6}} \right) = \sqrt 3 \cos x}$$ $${2\left( {\cos x \cdot \frac{{\sqrt 3 }}{2} - \sin x \cdot \frac{1}{2}} \right) = \sqrt 3 \cos x}$$ $${\sqrt 3 \cos x - \sin x - \sqrt 3 \cos x = 0}$$ $${\sin x = 0}$$ $${x = \pi n;\quad n \in Z}$$ Жауабы:$${\quad x = \pi n;\quad n \in Z}$$

Шешуі: $${\sin 3x - \sin x - \sin x = 0}$$ $${2\sin \frac{{3x - x}}{2}\cos \frac{{3x + x}}{2} - \sin x = 0}$$ $${2\sin x\cos 2x - \sin x = 0}$$ $${\sin x(2\cos 2x - 1) = 0}$$ $$1)\quad \sin x = 0$$ $${x = \pi k;\quad k \in Z}$$ $$2)\quad 2\cos 2x - 1 = 0$$ $${2\cos 2x = 1}$$ $${\cos 2x = \frac{1}{2}}$$ $${2x = \pm \arccos \frac{1}{2} + 2\pi n;\quad n \in Z}$$ $${2x = \pm \frac{\pi }{3} + 2\pi n;\quad n \in Z}$$ $${x = \pm \frac{\pi }{6} + \pi n;\quad n \in Z}$$ Жауабы: $$\quad \pi k;\quad \pm \frac{\pi }{6} + \pi n;\quad n,k \in Z$$

Шешуі:$${\tg \left( {x + 20^\circ } \right) + \tg \left( {90^\circ - \left( {x + 20^\circ } \right)} \right) = 2}$$ $${\tg \left( {x + 20^\circ } \right) + \ctg \left( {x + 20^\circ } \right) = 2}$$ $${\tg \left( {x + 20^\circ } \right) + \frac{1}{{\tg \left( {x + 20^\circ } \right)}} = 2}$$ $${\tg \left( {x + 20^\circ } \right) = t}$$ $${t + \frac{1}{t} = 2}$$ $${{t^2} - 2t + 1 = 0\quad t \ne 0}$$ $${{{(t - 1)}^2} = 0,\quad t - 1 = 0,\quad t = 1}$$ $${\tg \left( {x + 20^\circ } \right) = 1}$$ $${x + 20^\circ = \arctg 1 + \pi n;\quad n \in Z}$$ $${x + 20^\circ = 45^\circ + \pi n;\quad n \in Z}$$ $${x = 25^\circ + 180^\circ n;\quad n \in Z}$$ Жауабы:$\quad {x = 25^\circ + 180^\circ n;\quad n \in Z}$

Шешуі: $$\cos 2x\sin x - \cos 2x = 0$$ $$\cos 2x(\sin x - 1) = 0$$ $${1)\quad \cos 2x = 0}$$ $${2x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${x = \frac{\pi }{4} + \frac{{\pi n}}{2};\quad n \in Z}$$ Аралыққа тиісті шешімдері: $${90^\circ \lt 45^\circ + 90^\circ n \lt 180^\circ }$$ $${45^\circ \lt 90^\circ n \lt 135^\circ }$$ $${0,5 \lt n \lt 1,5}$$ $${n = 1}$$ $$x = 45^\circ + 90^\circ \cdot 1 = 135^\circ $$ $$2)\quad \sin x - 1 = 0$$ $$\sin x = 1$$ $$x = \frac{\pi }{2} + 2\pi k;\quad k \in Z$$ Аралыққа тиісті шешімдері: $${90^\circ \lt 90^\circ + 360^\circ k \lt 180^\circ }$$ $${0^\circ \lt 360^\circ k \lt 90^\circ }$$ $${0 \lt k \lt 0,25}$$ Жауабы: $135^\circ $

Шешуі: $${\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right) = 0}$$ Жауабы: $${\cos 2x = 0}$$ $${2x = \frac{\pi }{2} + \pi n;\quad n \in Z}$$ $${x = \frac{\pi }{4} + \frac{{\pi n}}{2};\quad n \in Z}$$ Жауабы: ${\quad \pi + \frac{{\pi n}}{2};\quad n \in Z.}$

Шешуі: $$2{\sin ^2}x - \sqrt 3 \cdot 2\sin x\cos x = 0$$ $$\sin x(2\sin x - 2\sqrt 3 \cos x) = 0$$ $${1)\quad \sin x = 0}$$ $${x = \pi k;\quad k \in Z}$$ Аралыққа тиісті шешімдері: $${0 \lt 180^\circ k \lt 90^\circ }$$ $${0 \lt x \lt 0,5}$$ $${2)\quad 2\sin x - 2\sqrt 3 \cos x = 0}$$ $${\frac{{2\sin x}}{{\cos x}} - \frac{{2\sqrt 3 \cos x}}{{\cos x}} = \frac{0}{{\cos x}}}$$ $${2\tg x - 2\sqrt 3 = 0}$$ $${2\tg x = 2\sqrt 3 }$$ $${\tg x = \sqrt 3 }$$ $${x = \arctg \sqrt 3 + \pi n;\quad n \in Z}$$ $${x = \frac{\pi }{3} + \pi n;\quad n \in Z}$$ Аралыққа тиісті шешімдері: $${0^\circ \lt 60^\circ + 180^\circ n \lt 90^\circ }$$ $${ - 60^\circ \lt 180^\circ n \lt 30^\circ }$$ $${ - 0,3 \lt n \lt 0,4}$$ $${n = 0}$$ $$x = 60^\circ + \pi \cdot 0 = 60^\circ $$ Жауабы: $60^\circ $