Тригонометриялық теңдеулер (Сканави (А) 8.1-8.15)

()

 +/-  - Есептің жауабын көрсету/көрсетпеу.

▲/▼ - Жауап орнын жасыру/шығару

   ×    - Сұрақты алып тастау.

Тригонометриялық теңдеулерді шешіңіз.

№ 8.1 Теңдеуді шешіңіз: $\cos 3x - \sin x = \sqrt 3 (\cos x - \sin 3x)$

Шешуі: $${\cos 3x - \sqrt 3 \cos x = \sin x - \sqrt 3 \sin 3x, \Leftrightarrow \dfrac{1}{2}\cos 3x + \dfrac{{\sqrt 3 }}{2}\sin 3x = }$$ $${ = \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x, \Leftrightarrow \cos 3x\cos \dfrac{\pi }{3} + \sin 3x\sin \dfrac{\pi }{3} = }$$ $${ = \cos \dfrac{\pi }{6}\cos x + \sin \dfrac{\pi }{6}\sin x, \Leftrightarrow \cos \left( {3x - \dfrac{\pi }{3}} \right) = \cos \left( {x - \dfrac{\pi }{6}} \right) \Leftrightarrow }$$ $$ \Leftrightarrow \cos \left( {3x - \dfrac{\pi }{3}} \right) - \cos \left( {x - \dfrac{\pi }{6}} \right) = 0, \Leftrightarrow - 2\sin \dfrac{{3x - \dfrac{\pi }{3} + x - \dfrac{\pi }{6}}}{2} \times \sin \dfrac{{3x - \dfrac{\pi }{3} - x + \dfrac{\pi }{6}}}{2} = 0$$ $$\sin \left( {2x - \dfrac{\pi }{4}} \right)\sin \left( {x - \dfrac{\pi }{{12}}} \right) = 0$$ $$1)\sin \left( {2x - \dfrac{\pi }{4}} \right) = 0,\quad 2{x_1} - \dfrac{\pi }{4} = \pi k,\quad {x_1} = \dfrac{\pi }{8} + \dfrac{{\pi k}}{2} = \dfrac{\pi }{8}(4k + 1)\quad k \in Z;$$ $$2)\sin \left( {x - \dfrac{\pi }{{12}}} \right) = 0,\quad {x_2} - \dfrac{\pi }{{12}} = \pi n,\quad {x_2} = \dfrac{\pi }{{12}} + \pi n = \dfrac{\pi }{{12}}(12n + 1),\quad n \in Z$$ $${x_1} = \dfrac{\pi }{8}(4k + 1),\,\,\,{x_2} = \dfrac{\pi }{{12}}(12n + 1),k,n \in Z$$

№ 8.2 Теңдеуді шешіңіз: $7 + 4\sin x\cos x + 1,5(\tg x + {{\rm ctg}} x) = 0$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{\sin x \ne 0,}\\{\cos x \ne 0}\end{array}} \right.$ $${7 + 4\sin x\cos x + 1,5\left( {\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}} \right) = 0 \Leftrightarrow }$$ $${ \Leftrightarrow 7 + 4\sin x\cos x + 1,5\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x\cos x}}} \right) = 0 \Leftrightarrow }$$ $${ \Leftrightarrow 7 + 4\sin x\cos x + \dfrac{{1,5}}{{\sin x\cos x}} = 0 \Leftrightarrow 7 + 2\sin 2x + \dfrac{3}{{\sin 2x}} = 0.}$$ $$2{\sin ^2}2x + 7\sin 2x + 3 = 0$$ $$\sin 2x = - 3,\emptyset \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin 2x = - \dfrac{1}{2}$$ $$2x = {( - 1)^k}\left( { - \dfrac{\pi }{6}} \right) + \pi k,\quad k \in Z,$$ $$x = {( - 1)^{k + 1}}\dfrac{\pi }{{12}} + \dfrac{{\pi k}}{2},\quad k \in Z$$

№ 8.3 Теңдеуді шешіңіз: $\dfrac{{4{{\rm ctg}} x}}{{1 + {{{{\rm ctg}} }^2}x}} + {\sin ^2}2x + 1 = 0$

Шешуі: ММЖ: $\sin x \ne 0$ $${\dfrac{{\dfrac{{4\cos x}}{{\sin x}}}}{{1 + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}} + {{\sin }^2}2x + 1 = 0, \Leftrightarrow {{\sin }^2}2x + 2\sin 2x + 1 = 0 \Leftrightarrow }$$ $${ \Leftrightarrow {{(\sin 2x + 1)}^2} = 0,\quad \sin 2x = - 1}$$ $$2x = - \dfrac{\pi }{2} + 2\pi k,\quad k \in Z,$$ $$x = - \dfrac{\pi }{4} + \pi k = \dfrac{\pi }{4}(4k - 1),\quad k \in Z.$$

№ 8.4 Теңдеуді шешіңіз: $\dfrac{{{{\sin }^2}2x - 4{{\sin }^2}x}}{{{{\sin }^2}2x + 4{{\sin }^2}x - 4}} + 1 = 2{\tg ^2}x$

Шешуі: ММЖ: $\cos x \ne 0$ $${\dfrac{{{{\sin }^2}2x - 4{{\sin }^2}x + {{\sin }^2}2x + 4{{\sin }^2}x - 4}}{{{{\sin }^2}2x + 4{{\sin }^2}x - 4}} = 2{{\tg }^2}x \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{2{{\sin }^2}2x - 4}}{{{{\sin }^2}2x + 4{{\sin }^2}x - 4}} = 2{{\tg }^2}x,\dfrac{{{{\sin }^2}2x - 2}}{{{{\sin }^2}2x + 4{{\sin }^2}x - 4}} = {{\tg }^2}x \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{{{\sin }^2}2x - 2}}{{{{\sin }^2}2x + 4{{\sin }^2}x - 4}} = \dfrac{{1 - \cos 2x}}{{1 + \cos 2x}} \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{1 - {{\cos }^2}2x - 2}}{{1 - {{\cos }^2}2x + 2 - 2\cos 2x - 4}} = \dfrac{{1 - \cos 2x}}{{1 + \cos 2x}} \Rightarrow {{\cos }^3}2x + {{\cos }^2}2x = 0,}$$ $${\cos ^2}2x(\cos 2x + 1) = 0.{\rm{ }}$$ $$1){\cos ^2}2x = 0,\quad \cos 2x = 0,\quad 2x = \dfrac{\pi }{2} + \pi k,\quad k \in Z$$ $${x_1} = \dfrac{\pi }{4} + \dfrac{{\pi k}}{2} = \dfrac{\pi }{4}(2k + 1)\quad k \in Z{\rm{; }}$$ $$2)\cos 2x + 1 = 0,\quad \cos 2x = - 1,\quad 2x = \pi + 2\pi n,\quad n \in Z$$ $${x_2} = \dfrac{\pi }{2} + \pi n = \dfrac{\pi }{2}(2n + 1),\quad n \in Z$$ $$x = \dfrac{\pi }{4} + \dfrac{{\pi m}}{2} = \dfrac{\pi }{4}(2m + 1),\quad m \in Z$$

№ 8.5 Теңдеуді шешіңіз: \(\sin z\left( {2\sin \left( {60^\circ - z} \right)\sin \left( {60^\circ + z} \right)} \right) = \dfrac{1}{4}\)

Шешуі: \[{\sin z\left( {2\sin \left( {60^\circ - z} \right)\sin \left( {60^\circ + z} \right)} \right) = \dfrac{1}{4}, \Leftrightarrow \sin z\left( {\cos 2z - \cos 120^\circ } \right) = \dfrac{1}{4} \Leftrightarrow }\] \[{ \Leftrightarrow 2\sin z\cos 2z + \sin z = \dfrac{1}{2}, \Leftrightarrow - \sin z + \sin 3z + \sin z = \dfrac{1}{2},\quad \sin 3z = \dfrac{1}{2}.}\] \[3z = {( - 1)^k} \cdot 30^\circ + 180^\circ k,\quad k \in Z,\quad z = {( - 1)^k} \cdot 10^\circ + 60^\circ k,\quad k \in Z\] \[z = {( - 1)^k} \cdot 10^\circ + 60^\circ k,\quad k \in Z.\]

№ 8.6 Теңдеуді шешіңіз: ${\cos ^{ - 2}}2t - {\sin ^{ - 2}}2t = \dfrac{8}{3}$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{\cos 2t \ne 0}\\{\sin 2t \ne 0}\end{array}} \right.$ $${\dfrac{1}{{{{\cos }^2}2t}} - \dfrac{1}{{{{\sin }^2}2t}} - \dfrac{8}{3} = 0,\,\,\,\dfrac{{{{\cos }^2}2t - {{\sin }^2}2t}}{{{{\sin }^2}2t{{\cos }^2}2t}} + \dfrac{8}{3} = 0, \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{\cos 4t}}{{{{\sin }^2}2t}} + \dfrac{2}{3} = 0,\dfrac{{\cos 4t}}{{1 - {{\cos }^2}4t}} + \dfrac{2}{3} = 0,\quad 2{{\cos }^2}4t - 3\cos 4t - 2 = 0.}$$ $$\cos 4t = 2,\emptyset ;\cos 4t = - \dfrac{1}{2},$$ $$4t = \pm \dfrac{2}{3}\pi + 2\pi k,\quad k \in Z,\quad t = \pm \dfrac{\pi }{6} + \dfrac{{\pi k}}{2},\,k \in Z.$$

№ 8.7 Теңдеуді шешіңіз: $\tg 3t - \tg t - 4\sin t = 0$

Шешуі: ММЖ: $\left\{ {\begin{array}{lllllllllllllll}{\cos 3t \ne 0}\\{\cos t \ne 0}\end{array}} \right.$ $$\boxed{\tg \alpha - \tg \beta = \dfrac{{\sin (\alpha - \beta )}}{{\cos \alpha \cos \beta }}},$$ \[\dfrac{{\sin 2t}}{{\cos 3t\cos t}} - 4\sin t = 0,\dfrac{{2\sin t\cos t}}{{\cos 3t\cos t}} - 4\sin t = 0, \Leftrightarrow 2\sin t \cdot \left( {\dfrac{{1 - 2\cos 3t}}{{\cos 3t}}} \right) = 0\] \[\sin t = 0,{t_1} = \pi k,k \in Z\] \[1 - 2\cos 3t = 0,\cos 3t = \dfrac{1}{2},3t = \pm \dfrac{\pi }{3} + 2\pi n,\] \[{t_2} = \pm \dfrac{\pi }{9} + \dfrac{{2\pi n}}{3},\quad n \in Z\]

№ 8.8 Теңдеуді шешіңіз: \({\cos ^{ - 1}}3t - 6\cos 3t = 4\sin 3t\)

Шешуі: ММЖ: $\cos 3t \ne 0$ \[1 - 6{\cos ^2}3t - 4\cos 3t\sin 3t = 0,\] \[{\cos ^2}3t + {\sin ^2}3t - 6{\cos ^2}3t - 4\cos 3t\sin 3t = 0\] \[5{\cos ^2}3t - {\sin ^2}3t + 4\cos 3t\sin 3t = 0\] \( - {\cos ^2}3t \ne 0,\) өрнегіне бөлеміз \[{\tg ^2}3t - 4\tg 3t - 5 = 0\] \[{(\tg 3t)_1} = - 1\] \[{(\tg 3t)_2} = 5\] \[3{t_1} = - \dfrac{\pi }{4} + \pi k,{t_1} = - \dfrac{\pi }{{12}} + \dfrac{{\pi k}}{3} = \dfrac{\pi }{{12}}(4k - 1),\,\,k \in Z\] \[3{t_2} = \arctg 5 + \pi n,{t_2} = \dfrac{{\arctg 5}}{3} + \dfrac{{\pi n}}{3},\,\,n \in Z\] \[{t_1} = \dfrac{\pi }{{12}}(4k - 1),{t_2} = \dfrac{{\arctg 5}}{3} + \dfrac{{\pi n}}{3},k,\,\,n \in Z.\]

№ 8.9 Теңдеуді шешіңіз: \[\ctg t - \sin t = 2{\sin ^2}\dfrac{t}{2}\]

Шешуі: ММЖ: \[\sin t \ne 0\] \[{\dfrac{{\cos t}}{{\sin t}} - \sin t = 1 - \cos t \Rightarrow \cos t - {{\sin }^2}t = \sin t - \sin t\cos t}\] \[{(\cos t + \sin t\cos t) - \left( {{{\sin }^2}t + \sin t} \right) = 0,\quad \cos t(1 + \sin t) - \sin t(1 + \sin t) = 0,}\] \[{(1 + \sin t)(\cos t - \sin t) = 0.}\] \[1)\,1 + \sin t = 0\] \[2)\cos t - \sin t = 0\] \[1)\sin t = - 1,\quad {t_1} = - \dfrac{\pi }{2} + 2\pi k = \dfrac{\pi }{2}(4k - 1)\quad k \in Z\] \[2)\cos t = \sin t \Leftrightarrow \tg t = 1,\quad {t_2} = \dfrac{\pi }{4} + \pi n = \dfrac{\pi }{4}(4n + 1),\quad n \in Z.\]

№ 8.10 Теңдеуді шешіңіз: $4\cos z\left( {\cos 2z + \cos {{120}^\circ }} \right) + 1 = 0,$

Шешуі: $$4\cos z\cos 2z - 2\cos z + 1 = 0 \Leftrightarrow $$ $$ \Leftrightarrow 2\cos z + 2\cos 3z - 2\cos z + 1 = 0,$$ $$\cos 3z = - \dfrac{1}{2},$$ $$3z = \pm \dfrac{2}{3}\pi + 2\pi k,$$ $${z = \pm \dfrac{2}{9}\pi + \dfrac{{2\pi k}}{3},\,\,k \in Z}$$

№ 8.11 Теңдеуді шешіңіз: $\sin \left( {\dfrac{\pi }{2} + 2x} \right)\tg 3x + \sin (\pi + 2x) - \sqrt 2 \cos 5x = 0$

Шешуі: ММЖ: $\sin 3x \ne 0$ $${\cos 2x\ctg 3x - \sin 2x - \sqrt 2 \cos 5x = 0 \Leftrightarrow }$$ $${ \Leftrightarrow \left( {\dfrac{{\cos 2x\cos 3x}}{{\sin 3x}} - \sin 2x} \right) - \sqrt 2 \cos 5x = 0}$$ $${\dfrac{{\cos 2x\cos 3x - \sin 2x\sin 3x}}{{\sin 3x}} - \sqrt 2 \cos 5x = 0 \Leftrightarrow }$$ $${ \Leftrightarrow \dfrac{{\cos 5x}}{{\sin 3x}} - \sqrt 2 \cos 5x = 0,\quad \dfrac{{\cos 5x(1 - \sqrt 2 \sin 3x)}}{{\sin 3x}} = 0.}$$ $$1)\,\cos 5x = 0,\quad 5x = \dfrac{\pi }{2} + \pi k,\quad {x_1} = \dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5} = \dfrac{\pi }{{10}}(2k + 1),\quad k \in Z,$$ $$2)\,1 - \sqrt 2 \sin 3x = 0,\quad \sin 3x = \dfrac{{\sqrt 2 }}{2},\quad 3x = {( - 1)^n}\dfrac{\pi }{4} + \pi n,\quad {x_2} = {( - 1)^n}\dfrac{\pi }{{12}} + \dfrac{{\pi n}}{3},\,n \in Z.$$

№ 8.12 Теңдеуді шешіңіз: $\sin x\cos 2x + \cos x\cos 4x = \sin \left( {\dfrac{\pi }{4} + 2x} \right)\sin \left( {\dfrac{\pi }{4} - 3x} \right)$

Шешуі: $${ - \sin x + \sin 3x + \cos 3x + \cos 5x = \cos 5x - \cos \left( {\dfrac{\pi }{2} - x} \right) \Leftrightarrow }$$ $${ \Leftrightarrow \sin 3x + \cos 3x = 0,\,\,\sin 3x = - \cos 3x,\,\,\tg 3x = - 1}$$ $$3x = - \dfrac{\pi }{4} + \pi n,\,\,x = - \dfrac{\pi }{{12}} + \dfrac{{\pi n}}{3} = \dfrac{\pi }{{12}}(4n - 1),\,\,n \in Z.$$

№ 8.13 Теңдеуді шешіңіз: $\sin 2x = {\cos ^4}\dfrac{x}{2} - {\sin ^4}\dfrac{x}{2}$

Шешуі: $$2\sin x\cos x - \left( {{{\cos }^2}\dfrac{x}{2} + {{\sin }^2}\dfrac{x}{2}} \right)\left( {{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}} \right) = 0,$$ $$2\sin x\cos x - \cos x = 0 \Leftrightarrow \cos x(2\sin x - 1) = 0$$ $$1)\,\cos x = 0,\,{x_1} = \dfrac{\pi }{2} + \pi n = \dfrac{\pi }{2}(2n + 1),\,n \in Z$$ $$2)2\sin x - 1 = 0,\,\sin x = \dfrac{1}{2},$$ $${x_2} = {( - 1)^k}\dfrac{\pi }{6} + \pi k,\,k \in Z$$ $${x_1} = \dfrac{\pi }{2}(2n + 1),{x_2} = {( - 1)^k}\dfrac{\pi }{6} + \pi k,\,\,n,k \in Z.$$

№ 8.14 Теңдеуді шешіңіз: $(1 + \cos 4x)\sin 2x = {\cos ^2}2x$

Шешуі: $$\left( {1 + 1 - 2{{\sin }^2}2x} \right)\sin 2x = 1 - {\sin ^2}2x,$$ $$2{\sin ^3}2x - {\sin ^2}2x - 2\sin 2x + 1 = 0,$$ $${\sin ^2}2x(2\sin 2x - 1) - (2\sin 2x - 1) = 0,$$ $$(2\sin 2x - 1)\left( {{{\sin }^2}2x - 1} \right) = 0.$$ $$1)\,2\sin 2x - 1 = 0,\,\,\sin 2x = \dfrac{1}{2},\,\,2x = {( - 1)^k}\dfrac{\pi }{6} + \pi k,\,\,{x_1} = {( - 1)^k}\dfrac{\pi }{{12}} + \dfrac{{\pi k}}{2},\,k \in Z,$$ $$2)\,{\sin ^2}2x - 1 = 0,\,\,\sin 2x = \pm 1,\,\,2x = \dfrac{\pi }{2} + \pi n,\,\,{x_2} = \dfrac{\pi }{4} + \dfrac{{\pi n}}{2} = \dfrac{\pi }{4}(2n + 1),\,n \in Z.$$

№ 8.15 Теңдеуді шешіңіз: ${\sin ^2}2z + {\sin ^2}3z + {\sin ^2}4z + {\sin ^2}5z = 2$

Шешуі: $${\dfrac{1}{2}(1 - \cos 4z) + \dfrac{1}{2}(1 - \cos 6z) + \dfrac{1}{2}(1 - \cos 8z) + \dfrac{1}{2}(1 - \cos 10z) = 2}$$ $${(\cos 4z + \cos 6z) + (\cos 8z + \cos 10z) = 0, \Leftrightarrow 2\cos 5z\cos z + 2\cos 9z\cos z = 0}$$ $$2\cos z(\cos 5z + \cos 9z) = 0$$ $$1)\,\cos z = 0,\,{z_1} = \dfrac{\pi }{2} + \pi k = \dfrac{\pi }{2}(2k + 1),\,k \in Z,$$ $$2)\,\cos 5z + \cos 9z = 0,\,\,\,\,\,2\cos 7z\cos 2z = 0,\,\,\,\,\cos 7z = 0,\,\,\,\,7z = \dfrac{\pi }{2} + \pi n,$$ $${z_2} = \dfrac{\pi }{{14}} + \dfrac{{\pi n}}{7} = \dfrac{\pi }{{14}}(2n + 1),\,n \in Z;$$ $$3)\,\cos 2z = 0,2z = \dfrac{\pi }{2} + \pi m,\,\,{z_3} = \dfrac{\pi }{4} + \dfrac{{\pi m}}{2} = \dfrac{\pi }{4}(2m + 1),m \in Z;$$ $${z_1} \subset {z_2}$$ Жауабы ${z_1} = \dfrac{\pi }{{14}}(2n + 1),\,\,{z_2} = \dfrac{\pi }{4}(2m + 1),\,\,\,\,n,m \in Z.$

 

Жазба сіз үшін қаншалықты қажет болды?

Жұлдызшаның үстінен басыңыз!

Сіз бұл жазбаны қажетті деп таптыңыз...

Әлеуметтік желіде бөлісіңіз!

Бұл жазбаның сіз үшін қажетті болмағаны өкінішті!

Жазбамызды жақсартайық!

Жазбаны жақсартуға қандай ұсыныс айтар едіңіз?

Осы тақырыптағы посттар

Пікір қалдыру

Сіздің электронды почтаңыз жарияланбайды, Міндетті жолдарды толтырып шығыңыз.