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3cos-x-4sin-x-tan-2-x-locate-x-




Question Number 37155 by ajfour last updated on 09/Jun/18
3cos x−4sin x=tan^2 x  locate x.
$$\mathrm{3cos}\:{x}−\mathrm{4sin}\:{x}=\mathrm{tan}\:^{\mathrm{2}} {x} \\ $$$${locate}\:{x}. \\ $$
Commented by behi83417@gmail.com last updated on 09/Jun/18
c=cosx,s=sinx  3c−4s=(s^2 /c^2 )⇒3c^3 −4sc^2 =1−c^2   ⇒4sc^2 =3c^3 +c^2 −1⇒16s^2 c^4 =(3c^3 +c^2 −1)^2   ⇒25c^6 +6c^5 −15c^4 −6c^3 −2c^2 +1=0  ⇒c=0.41,0.85  cosx=0.41⇒x=360k±65.8^•   cosx=0.85⇒x=360l±31.8^•  .■
$${c}={cosx},{s}={sinx} \\ $$$$\mathrm{3}{c}−\mathrm{4}{s}=\frac{{s}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\Rightarrow\mathrm{3}{c}^{\mathrm{3}} −\mathrm{4}{sc}^{\mathrm{2}} =\mathrm{1}−{c}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{sc}^{\mathrm{2}} =\mathrm{3}{c}^{\mathrm{3}} +{c}^{\mathrm{2}} −\mathrm{1}\Rightarrow\mathrm{16}{s}^{\mathrm{2}} {c}^{\mathrm{4}} =\left(\mathrm{3}{c}^{\mathrm{3}} +{c}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{25}{c}^{\mathrm{6}} +\mathrm{6}{c}^{\mathrm{5}} −\mathrm{15}{c}^{\mathrm{4}} −\mathrm{6}{c}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{c}=\mathrm{0}.\mathrm{41},\mathrm{0}.\mathrm{85} \\ $$$$\boldsymbol{{cosx}}=\mathrm{0}.\mathrm{41}\Rightarrow\boldsymbol{{x}}=\mathrm{360}\boldsymbol{{k}}\pm\mathrm{65}.\mathrm{8}^{\bullet} \\ $$$$\boldsymbol{{cosx}}=\mathrm{0}.\mathrm{85}\Rightarrow\boldsymbol{{x}}=\mathrm{360}\boldsymbol{{l}}\pm\mathrm{31}.\mathrm{8}^{\bullet} \:.\blacksquare \\ $$$$ \\ $$
Commented by ajfour last updated on 10/Jun/18
thank you Sir. (is calculator must ?)
$${thank}\:{you}\:{Sir}.\:\left({is}\:{calculator}\:{must}\:?\right) \\ $$
Commented by MJS last updated on 10/Jun/18
I think so  I tried to set cos x=(√(1−sin^2  x)) and vice versa  and it always leads to a polynome of degree 6
$$\mathrm{I}\:\mathrm{think}\:\mathrm{so} \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{set}\:\mathrm{cos}\:{x}=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}\:\mathrm{and}\:\mathrm{vice}\:\mathrm{versa} \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{always}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{6} \\ $$
Commented by ajfour last updated on 10/Jun/18
Sir please attempt Q.37166 also.
$${Sir}\:{please}\:{attempt}\:{Q}.\mathrm{37166}\:{also}. \\ $$

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