Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 174790 by MathsFan last updated on 11/Aug/22

Commented by MathsFan last updated on 11/Aug/22

any help?

$$\mathrm{any}\:\mathrm{help}? \\ $$

Answered by cortano1 last updated on 11/Aug/22

Answered by Ar Brandon last updated on 11/Aug/22

6. 3(2sinx−cosx)=2(sinx−3cosx) 6sinx−3cosx=2sinx−6cosx 4sinx=−3cosx tanx=−(3/4)⇒x=tan^(−1) (−(3/4)) PV ⇒x=π−arctan((3/4)) ∨ x=−arctan((3/4))

$$\mathrm{6}. \\ $$$$\mathrm{3}\left(\mathrm{2sin}{x}−\mathrm{cos}{x}\right)=\mathrm{2}\left(\mathrm{sin}{x}−\mathrm{3cos}{x}\right) \\ $$$$\mathrm{6sin}{x}−\mathrm{3cos}{x}=\mathrm{2sin}{x}−\mathrm{6cos}{x} \\ $$$$\mathrm{4sin}{x}=−\mathrm{3cos}{x} \\ $$$$\mathrm{tan}{x}=−\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} \left(−\frac{\mathrm{3}}{\mathrm{4}}\right)\:\:\:\:\:{PV} \\ $$$$\Rightarrow{x}=\pi−\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:\vee\:{x}=−\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$

Answered by Ar Brandon last updated on 11/Aug/22

7. 2sinxtanx+3=0 ⇒((2sin^2 x)/(cosx))+3=0 ⇒2(1−cos^2 x)+3cosx=0 ⇒2cos^2 x−3cosx−2=0 ⇒(2cosx+1)(cosx−2)=0 ⇒2cosx+1=0 (−1≤cosx≤1) ⇒cosx=−(1/2), PV=(π/3) ⇒x=π−(π/3)=((2π)/3), x=π+(π/3)=((4π)/3)

$$\mathrm{7}. \\ $$$$\mathrm{2sin}{x}\mathrm{tan}{x}+\mathrm{3}=\mathrm{0}\:\Rightarrow\frac{\mathrm{2sin}^{\mathrm{2}} {x}}{\mathrm{cos}{x}}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right)+\mathrm{3cos}{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2cos}^{\mathrm{2}} {x}−\mathrm{3cos}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2cos}{x}+\mathrm{1}\right)\left(\mathrm{cos}{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2cos}{x}+\mathrm{1}=\mathrm{0}\:\:\left(−\mathrm{1}\leqslant\mathrm{cos}{x}\leqslant\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{cos}{x}=−\frac{\mathrm{1}}{\mathrm{2}},\:{PV}=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\pi−\frac{\pi}{\mathrm{3}}=\frac{\mathrm{2}\pi}{\mathrm{3}},\:{x}=\pi+\frac{\pi}{\mathrm{3}}=\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com