Question Number 77027 by mathocean1 last updated on 02/Jan/20
![Please help me to solve it in [−π;0] cos2x+cosx+1=sin3x+sin2x+sinx Explain details if possible.](https://www.tinkutara.com/question/Q77027.png)
$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\left[−\pi;\mathrm{0}\right] \\ $$$$\mathrm{cos2}{x}+\mathrm{cos}{x}+\mathrm{1}=\mathrm{sin3}{x}+\mathrm{sin2}{x}+\mathrm{sin}{x} \\ $$$${E}\mathrm{xplain}\:\mathrm{details}\:\mathrm{if}\:\mathrm{possible}. \\ $$$$ \\ $$
Answered by mr W last updated on 02/Jan/20
![cos2x+cosx+1=sin3x+sin2x+sinx 2 cos^2 x+cosx=3 sin x−4 sin^3 x+2 sin x cos x+sinx (2 cos x+1)cosx=2 sin x cos x (2cos x+1) (2 cos x+1)(2 sin x−1)cosx=0 ⇒cos x=0 ⇒x=−(π/2) ⇒cos x=−(1/2) ⇒x=−((2π)/3) ⇒sin x=(1/2) ⇒no solution within [−π,0] ⇒solutions are: x=−((2π)/3), −(π/2)](https://www.tinkutara.com/question/Q77029.png)
$$\mathrm{cos2}{x}+\mathrm{cos}{x}+\mathrm{1}=\mathrm{sin3}{x}+\mathrm{sin2}{x}+\mathrm{sin}{x} \\ $$$$\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{cos}{x}=\mathrm{3}\:\mathrm{sin}\:{x}−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:{x}+\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}+\mathrm{sin}{x} \\ $$$$\left(\mathrm{2}\:\mathrm{cos}\:{x}+\mathrm{1}\right)\mathrm{cos}{x}=\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\left(\mathrm{2cos}\:{x}+\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\:\mathrm{cos}\:{x}+\mathrm{1}\right)\left(\mathrm{2}\:\mathrm{sin}\:{x}−\mathrm{1}\right)\mathrm{cos}{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\mathrm{0}\:\Rightarrow{x}=−\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{x}=−\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{no}\:{solution}\:{within}\:\left[−\pi,\mathrm{0}\right] \\ $$$$\Rightarrow{solutions}\:{are}:\:{x}=−\frac{\mathrm{2}\pi}{\mathrm{3}},\:−\frac{\pi}{\mathrm{2}} \\ $$
Commented by mathocean1 last updated on 02/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$