Сканави (А) 8.1-8.15

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№ 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: \[{\sin z\left( {2\sin \left( {60^\circ - z} \right)\sin \left( {60^\circ + z} \right)} \right) = \dfrac{1}{4}, \Leftrightarrow \sin z\left( {\cos 2z - \cos 120^\circ } \right) = \dfrac{1}{4} \Leftrightarrow }\] \[{ \Leftrightarrow 2\sin z\cos 2z + \sin z = \dfrac{1}{2}, \Leftrightarrow - \sin z + \sin 3z + \sin z = \dfrac{1}{2},\quad \sin 3z = \dfrac{1}{2}.}\] \[3z = {( - 1)^k} \cdot 30^\circ + 180^\circ k,\quad k \in Z,\quad z = {( - 1)^k} \cdot 10^\circ + 60^\circ k,\quad k \in Z\$ O\] \[z = {( - 1)^k} \cdot 10^\circ + 60^\circ k,\quad k \in Z.\]

№ 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 ${z_1} = \dfrac{\pi }{{14}}(2n + 1),\,\,{z_2} = \dfrac{\pi }{4}(2m + 1),\,\,\,\,n,m \in Z.$

 
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