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2-sin-x-cos-x-2-




Question Number 189952 by SAMIRA last updated on 25/Mar/23
(√2) sin(x) +cos(x)= −(√2)
$$\sqrt{\mathrm{2}}\:\mathrm{sin}\left(\mathrm{x}\right)\:+\mathrm{cos}\left(\mathrm{x}\right)=\:−\sqrt{\mathrm{2}}\: \\ $$
Answered by cortano12 last updated on 25/Mar/23
 (√3) (((√2)/( (√3))) sin x+(1/( (√3))) cos x)=−(√2)   (√3) (sin x cos θ+cos x sin θ)=−(√2)    sin (x+θ)=−((√2)/( (√3))) =sin (180°+α)   where  { ((θ=35°)),((α=55°)) :}
$$\:\sqrt{\mathrm{3}}\:\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}\:\mathrm{x}\right)=−\sqrt{\mathrm{2}} \\ $$$$\:\sqrt{\mathrm{3}}\:\left(\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\theta+\mathrm{cos}\:\mathrm{x}\:\mathrm{sin}\:\theta\right)=−\sqrt{\mathrm{2}} \\ $$$$\:\:\mathrm{sin}\:\left(\mathrm{x}+\theta\right)=−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:=\mathrm{sin}\:\left(\mathrm{180}°+\alpha\right) \\ $$$$\:\mathrm{where}\:\begin{cases}{\theta=\mathrm{35}°}\\{\alpha=\mathrm{55}°}\end{cases} \\ $$
Answered by mehdee42 last updated on 25/Mar/23
let  y=tan(x/2)⇒(√2)×((2y)/(1+y^2 ))+((1−y^2 )/(1+y^2 ))=−(√2)  (1−(√2))y^2 −2(√2)y−(√2)−1=0  y=−((√2)+1)^2 =tanα⇒x=2kπ+2α  y=−1=tan(−(π/4))⇒x=2kπ−(π/2)
$${let}\:\:{y}={tan}\frac{{x}}{\mathrm{2}}\Rightarrow\sqrt{\mathrm{2}}×\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }+\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }=−\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{y}−\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{0} \\ $$$${y}=−\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} ={tan}\alpha\Rightarrow{x}=\mathrm{2}{k}\pi+\mathrm{2}\alpha \\ $$$${y}=−\mathrm{1}={tan}\left(−\frac{\pi}{\mathrm{4}}\right)\Rightarrow{x}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}} \\ $$

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