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Answers

Answered by rishu6845
3

Answer:

\boxed{\huge{\pink{5}}}

Step-by-step explanation:

\bold{Given}\longrightarrow \\ cos3x + cos2x = sin \dfrac{3x}{2}  + sin \dfrac{x}{2}

\bold{To \: find }\longrightarrow\\ the \: number \: of \: roots \: of \: the \: given \: equation \\ when \:  \\ 0 \leqslant x \leqslant 2\pi

\bold{Concept \: used}\longrightarrow \\ 1) \: cos \alpha  + cos \beta  = 2cos( \dfrac{ \alpha  +  \beta }{2} )cos( \dfrac{ \alpha  -  \beta }{2} ) \\ 2)sin \alpha  + sin \beta  = 2sin( \dfrac{ \alpha  +  \beta }{2} )cos( \dfrac{ \alpha  -  \beta }{2} ) \\ 3)general \: value \: of \: cos \\ x = 2n\pi \pm \alpha

\bold{Soution }\longrightarrow\\ cos3x + cos2x = sin \dfrac{3x}{2 }  + sin \dfrac{x}{2}  \\  =  > 2cos( \dfrac{3x + 2x}{2} )cos( \dfrac{3x - 2x}{2} ) = 2sin( \dfrac{ \dfrac{3x}{2}  +  \dfrac{x}{2} }{2} )cos( \dfrac{ \dfrac{3x}{2}  -  \dfrac{x}{2} }{2} ) \\  =  > 2cos( \dfrac{5x}{2} ) \: cos( \dfrac{x}{2})  = 2sinx \: cos( \dfrac{x}{2} ) \\ 2 \: cos \dfrac{x}{2}  \: is \: cancel \: out \: from \: each \: side \\  =  > cos \dfrac{5x}{2}  \:  =  \: sinx \\  =  > cos \dfrac{5x}{2}  \:  =  \: cos( \dfrac{\pi}{2}  - x) \\  =  > cos \dfrac{5x}{2}  = cos( \: 2n\pi \:  \pm \: ( \dfrac{\pi}{2}  - x) \: ) \\  =  >  \dfrac{5x}{2}  = 2n\pi \:  \pm \: ( \dfrac{\pi}{2}  - x) \\ taking \:  +  \: sign \: we \: get \\   =  > \dfrac{5x}{2}  = 2n\pi +  \dfrac{\pi}{2}  - x \\  =  >  \dfrac{5x}{2}  + x = 2n\pi  +  \dfrac{\pi}{2} \\  =  >  \dfrac{7x}{2}  = 2n\pi +  \dfrac{\pi}{2 }  \\  =  > x =  \dfrac{2}{7} \:  \:  2n\pi +  \dfrac{2}{7}   \:  \: \dfrac{\pi}{2}  \\  =  > x =  \dfrac{4n\pi}{7}  +  \dfrac{\pi}{7}  \\  =  > x = (4n + 1) \dfrac{\pi}{7}

putting \: n \:  = 0 \: in \: it

x =  \dfrac{\pi}{7}  \\ 0  \leqslant  \dfrac{\pi}{7}  \leqslant 2\pi

putting \: n = 1 \: in \: it \: we \: get \\ x =  \dfrac{5\pi}{7}  \\ 0 \leqslant  \dfrac{5\pi}{7}  \leqslant 2\pi

putting \: n = 2 \: we \: get \\ x =  \dfrac{9\pi}{7}  \\ 0 \leqslant  \dfrac{9\pi}{7}  \leqslant 2\pi \\ putting \: n = 3 \\  x =  \dfrac{13\pi}{7 }  \\ 0 \leqslant  \dfrac{13\pi}{7}  \leqslant 2\pi \\ putting \: n = 4 \: we \:g et \\ x =  \dfrac{17\pi}{7}  \\   2\pi <  \dfrac{17\pi}{7}  \\ so \: it \: is \: not \: in \: given \: range \:  \\ it \: means \: all \: values \: of \: x \: for \: which  \\ \: n > 3 \: is \: not \: in \: given \: range \\ now \: taking \: negative \: sign

 \dfrac{5x}{2}  = 2n\pi - ( \dfrac{\pi}{2}  - x) \\  =  >  \dfrac{5x}{2}  = 2n\pi -  \dfrac{\pi}{2}  + x \\  =  >  \dfrac{5x}{2}  - x = 2n\pi -  \dfrac{\pi}{2}  \\  =  >  \dfrac{3x}{2}  = 2n\pi -  \dfrac{\pi}{2}  \\  =  > x =  \dfrac{4n\pi}{3}  -  \dfrac{\pi}{3}  \\  =  > x = (4n - 1) \dfrac{\pi}{3}

putting \: n = 0 \: we \: get \\ x =  -  \dfrac{\pi}{3}  < 0 \\ so \: it \: is \: not \: in \: given \: range \\ putting \: n = 1 \: we \: get \\ x = \pi \\ it \: is\:in \: given \: range \\ putting \: n \:  = 2 \: we \: get \\ x =  \dfrac{7\pi}{3}  > 2\pi \\ so \: it \: i s\: not \: in \: given \: range \\ so \: values \: for \: which \: n > 1 \: \\  x \: does \: not \: lie \: i n\: given \: range

now

values \: of \: x \: which \:satisfy \: given \: equation  \\ \: and \:  lie \: in \: given \: range \: are \:  \\ x =  \dfrac{\pi}{7}, \:   \frac{5\pi}{7} , \:  \dfrac{9\pi}{7} , \:  \dfrac{13\pi}{7} , \: \pi \\ these \:  values\: are \: five \:  \\ so \: aswer \: is \:5

Answered by shubhamugale2005
0

Answer:

Step-by-step explanation:

\begin{gathered}\bold{Given}\longrightarrow \\ cos3x + cos2x = sin \dfrac{3x}{2} + sin \dfrac{x}{2} \end{gathered}

Given⟶

cos3x+cos2x=sin

2

3x

+sin

2

x

\begin{gathered}\bold{To \: find }\longrightarrow\\ the \: number \: of \: roots \: of \: the \: given \: equation \\ when \: \\ 0 \leqslant x \leqslant 2\pi\end{gathered}

Tofind⟶

thenumberofrootsofthegivenequation

when

0⩽x⩽2π

\begin{gathered}\bold{Concept \: used}\longrightarrow \\ 1) \: cos \alpha + cos \beta = 2cos( \dfrac{ \alpha + \beta }{2} )cos( \dfrac{ \alpha - \beta }{2} ) \\ 2)sin \alpha + sin \beta = 2sin( \dfrac{ \alpha + \beta }{2} )cos( \dfrac{ \alpha - \beta }{2} ) \\ 3)general \: value \: of \: cos \\ x = 2n\pi \pm \alpha \end{gathered}

Conceptused⟶

1)cosα+cosβ=2cos(

2

α+β

)cos(

2

α−β

)

2)sinα+sinβ=2sin(

2

α+β

)cos(

2

α−β

)

3)generalvalueofcos

x=2nπ±α

\begin{gathered}\bold{Soution }\longrightarrow\\ cos3x + cos2x = sin \dfrac{3x}{2 } + sin \dfrac{x}{2} \\ = > 2cos( \dfrac{3x + 2x}{2} )cos( \dfrac{3x - 2x}{2} ) = 2sin( \dfrac{ \dfrac{3x}{2} + \dfrac{x}{2} }{2} )cos( \dfrac{ \dfrac{3x}{2} - \dfrac{x}{2} }{2} ) \\ = > 2cos( \dfrac{5x}{2} ) \: cos( \dfrac{x}{2}) = 2sinx \: cos( \dfrac{x}{2} ) \\ 2 \: cos \dfrac{x}{2} \: is \: cancel \: out \: from \: each \: side \\ = > cos \dfrac{5x}{2} \: = \: sinx \\ = > cos \dfrac{5x}{2} \: = \: cos( \dfrac{\pi}{2} - x) \\ = > cos \dfrac{5x}{2} = cos( \: 2n\pi \: \pm \: ( \dfrac{\pi}{2} - x) \: ) \\ = > \dfrac{5x}{2} = 2n\pi \: \pm \: ( \dfrac{\pi}{2} - x) \\ taking \: + \: sign \: we \: get \\ = > \dfrac{5x}{2} = 2n\pi + \dfrac{\pi}{2} - x \\ = > \dfrac{5x}{2} + x = 2n\pi + \dfrac{\pi}{2} \\ = > \dfrac{7x}{2} = 2n\pi + \dfrac{\pi}{2 } \\ = > x = \dfrac{2}{7} \: \: 2n\pi + \dfrac{2}{7} \: \: \dfrac{\pi}{2} \\ = > x = \dfrac{4n\pi}{7} + \dfrac{\pi}{7} \\ = > x = (4n + 1) \dfrac{\pi}{7} \end{gathered}

Soution⟶

cos3x+cos2x=sin

2

3x

+sin

2

x

=>2cos(

2

3x+2x

)cos(

2

3x−2x

)=2sin(

2

2

3x

+

2

x

)cos(

2

2

3x

2

x

)

=>2cos(

2

5x

)cos(

2

x

)=2sinxcos(

2

x

)

2cos

2

x

iscanceloutfromeachside

=>cos

2

5x

=sinx

=>cos

2

5x

=cos(

2

π

−x)

=>cos

2

5x

=cos(2nπ±(

2

π

−x))

=>

2

5x

=2nπ±(

2

π

−x)

taking+signweget

=>

2

5x

=2nπ+

2

π

−x

=>

2

5x

+x=2nπ+

2

π

=>

2

7x

=2nπ+

2

π

=>x=

7

2

2nπ+

7

2

2

π

=>x=

7

4nπ

+

7

π

=>x=(4n+1)

7

π

putting \: n \: = 0 \: in \: itputtingn=0init

\begin{gathered}x = \dfrac{\pi}{7} \\ 0 \leqslant \dfrac{\pi}{7} \leqslant 2\pi\end{gathered}

x=

7

π

0⩽

7

π

⩽2π

\begin{gathered}putting \: n = 1 \: in \: it \: we \: get \\ x = \dfrac{5\pi}{7} \\ 0 \leqslant \dfrac{5\pi}{7} \leqslant 2\pi\end{gathered}

puttingn=1initweget

x=

7

0⩽

7

⩽2π

\begin{gathered}putting \: n = 2 \: we \: get \\ x = \dfrac{9\pi}{7} \\ 0 \leqslant \dfrac{9\pi}{7} \leqslant 2\pi \\ putting \: n = 3 \\ x = \dfrac{13\pi}{7 } \\ 0 \leqslant \dfrac{13\pi}{7} \leqslant 2\pi \\ putting \: n = 4 \: we \:g et \\ x = \dfrac{17\pi}{7} \\ 2\pi < \dfrac{17\pi}{7} \\ so \: it \: is \: not \: in \: given \: range \: \\ it \: means \: all \: values \: of \: x \: for \: which \\ \: n > 3 \: is \: not \: in \: given \: range \\ now \: taking \: negative \: sign \end{gathered}

puttingn=2weget

x=

7

0⩽

7

⩽2π

puttingn=3

x=

7

13π

0⩽

7

13π

⩽2π

puttingn=4weget

x=

7

17π

2π<

7

17π

soitisnotingivenrange

itmeansallvaluesofxforwhich

n>3isnotingivenrange

nowtakingnegativesign

\begin{gathered} \dfrac{5x}{2} = 2n\pi - ( \dfrac{\pi}{2} - x) \\ = > \dfrac{5x}{2} = 2n\pi - \dfrac{\pi}{2} + x \\ = > \dfrac{5x}{2} - x = 2n\pi - \dfrac{\pi}{2} \\ = > \dfrac{3x}{2} = 2n\pi - \dfrac{\pi}{2} \\ = > x = \dfrac{4n\pi}{3} - \dfrac{\pi}{3} \\ = > x = (4n - 1) \dfrac{\pi}{3} \end{gathered}

2

5x

=2nπ−(

2

π

−x)

=>

2

5x

=2nπ−

2

π

+x

=>

2

5x

−x=2nπ−

2

π

=>

2

3x

=2nπ−

2

π

=>x=

3

4nπ

3

π

=>x=(4n−1)

3

π

\begin{gathered}putting \: n = 0 \: we \: get \\ x = - \dfrac{\pi}{3} < 0 \\ so \: it \: is \: not \: in \: given \: range \\ putting \: n = 1 \: we \: get \\ x = \pi \\ it \: is\:in \: given \: range \\ putting \: n \: = 2 \: we \: get \\ x = \dfrac{7\pi}{3} > 2\pi \\ so \: it \: i s\: not \: in \: given \: range \\ so \: values \: for \: which \: n > 1 \: \\ x \: does \: not \: lie \: i n\: given \: range\end{gathered}

puttingn=0weget

x=−

3

π

<0

soitisnotingivenrange

puttingn=1weget

x=π

itisingivenrange

puttingn=2weget

x=

3

>2π

soitisnotingivenrange

sovaluesforwhichn>1

xdoesnotlieingivenrange

nownow

\begin{gathered}values \: of \: x \: which \:satisfy \: given \: equation \\ \: and \: lie \: in \: given \: range \: are \: \\ x = \dfrac{\pi}{7}, \: \frac{5\pi}{7} , \: \dfrac{9\pi}{7} , \: \dfrac{13\pi}{7} , \: \pi \\ these \: values\: are \: five \: \\ so \: aswer \: is \:5\end{gathered}

valuesofxwhichsatisfygivenequation

andlieingivenrangeare

x=

7

π

,

7

,

7

,

7

13π

thesevaluesarefive

soasweris5

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