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Answers
Answer:
Step-by-step explanation:
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
5π
0⩽
7
5π
⩽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
9π
0⩽
7
9π
⩽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
7π
>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
5π
,
7
9π
,
7
13π
,π
thesevaluesarefive
soasweris5