Question 1

M109) solve

cos²x + 2 cosx + 1 = 0

Question 2

Answer:

Explanation:

Question 3

$\: \: \: \: \: \: \: \: \: \huge \color{green}{ \fbox \colorbox{black}{ \fbox{ \green {Answer} \: }}}$

$\longrightarrow \sf \color{teal}{ \cos {}^(2) (x) + 2 \cos(x) + 1} = 0$

$\longrightarrow \sf \color{teal}{ \cos {}^(2) (x) + \cos(x) + \cos(x) + 1} = 0$

$\longrightarrow \sf \color{teal}{ \cos {}^{} (x) ( \cos(x) + 1)+ 1(\cos(x) + 1} )= 0$

$\longrightarrow \sf \color{teal}{ ( \cos(x) + 1)(\cos(x) + 1} )= 0$

$\longrightarrow \sf \color{teal}{ ( (\cos(x) + 1} {}^{} ) {}^(2) = 0$

$\longrightarrow \sf \color{teal}{ ( (\cos(x) + 1} {}^{} ) {}^{} = 0$

$\longrightarrow \sf \color{teal}{ \cos(x) {}^{} }= - 1$

$\longrightarrow \sf \color{teal}{ x{}^{} }= \cos {}^( - 1)( - 1)$

$\longrightarrow \sf \color{teal}{ x = (2n + 1) \pi \: \: \: \: rad}$

where,
${ n \in \{whole \:\; number\}}$

Principle value -

$\longrightarrow \sf \color{teal}{ x} {}^{} {}^{} = \cos {}^( - 1) ( - 1)$

$\longrightarrow \sf \color{teal}{ x} {}^{} {}^{} = (2n + 1) \pi \: \: \: \: rad \:$

put n = 0

$\longrightarrow \sf \color{teal}{ x} {}^{} {}^{} = (2(0) + 1) \pi \: \: \: \: rad \:$

$\longrightarrow \sf \color{teal}{ x} {}^{} {}^{} = (0 + 1) \pi \: \: \: \: rad \:$

$\longrightarrow \sf \color{teal}{ x} {}^{} {}^{} = \pi \: \: \: \: rad \: = 180 \degree$

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