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Question Number 164671 by mnjuly1970 last updated on 20/Jan/22

          solve         cos^( 3) (x) + sin^( 2) (x) = (7/8)            adopted from youtube ...

$$ \\ $$$$\:\:\:\:\:\:\:\:{solve}\: \\ $$$$\:\:\:\:\:\:{cos}^{\:\mathrm{3}} \left({x}\right)\:+\:{sin}^{\:\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{7}}{\mathrm{8}}\: \\ $$$$\:\:\:\:\:\:\:\:\:{adopted}\:{from}\:{youtube}\:... \\ $$$$ \\ $$

Commented by MJS_new last updated on 20/Jan/22

for 0≤x<2π I get  x∈{(π/5), (π/3), ((3π)/5), ((7π)/5), ((5π)/3), ((9π)/5)}

$$\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{I}\:\mathrm{get} \\ $$$${x}\in\left\{\frac{\pi}{\mathrm{5}},\:\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{3}\pi}{\mathrm{5}},\:\frac{\mathrm{7}\pi}{\mathrm{5}},\:\frac{\mathrm{5}\pi}{\mathrm{3}},\:\frac{\mathrm{9}\pi}{\mathrm{5}}\right\} \\ $$

Answered by bobhans last updated on 20/Jan/22

 cos^3 x+1−cos^2 x−(7/8)=0    cos^3 x−cos^2 x+(1/8)=0   8cos^3 x−8cos^2 x+1 = 0   (2cos x−1)(4cos^2 x−2cos x−1)=0   (1)cos x=(1/2) ⇒x=± (π/3)+2kπ  (2)4cos^2 x−2cos x−1=0         cos^2 x−(1/2)cos x−(1/4)=0         (cos x−(1/4))^2 −(5/(16))=0        cos x−(1/4) = ± ((√5)/4)        cos x= ((1±(√5))/4)

$$\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}+\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\frac{\mathrm{7}}{\mathrm{8}}=\mathrm{0}\: \\ $$$$\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\:\mathrm{8cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{8cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\left(\mathrm{2cos}\:\mathrm{x}−\mathrm{1}\right)\left(\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2cos}\:\mathrm{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\left(\mathrm{1}\right)\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{x}=\pm\:\frac{\pi}{\mathrm{3}}+\mathrm{2k}\pi \\ $$$$\left(\mathrm{2}\right)\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2cos}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{cos}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{16}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\:=\:\pm\:\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\mathrm{x}=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}}\:\: \\ $$

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