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Question Number 8311 by lepan last updated on 07/Oct/16

Solve the equation 6cos2a−5sin2a=1.8  for0°≤a≤180°.

$${Solve}\:{the}\:{equation}\:\mathrm{6}{cos}\mathrm{2}{a}−\mathrm{5}{sin}\mathrm{2}{a}=\mathrm{1}.\mathrm{8} \\ $$$${for}\mathrm{0}°\leqslant{a}\leqslant\mathrm{180}°. \\ $$

Commented by 123456 last updated on 07/Oct/16

6cos2a−5sin2a=1.8  cos^2 2a+sin^2 2a=1  cos^2 2a=1−sin^2 2a  sin2a=x  ±6(√((1−x^2 )))=1.8+5x  36(1−x^2 )=(1.8+5x)^2

$$\mathrm{6cos2}{a}−\mathrm{5sin2}{a}=\mathrm{1}.\mathrm{8} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{a}+\mathrm{sin}^{\mathrm{2}} \mathrm{2}{a}=\mathrm{1} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{a}=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{2}{a} \\ $$$$\mathrm{sin2}{a}={x} \\ $$$$\pm\mathrm{6}\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}=\mathrm{1}.\mathrm{8}+\mathrm{5}{x} \\ $$$$\mathrm{36}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\left(\mathrm{1}.\mathrm{8}+\mathrm{5}{x}\right)^{\mathrm{2}} \\ $$

Answered by prakash jain last updated on 08/Oct/16

6cos 2a−5sin 2a=1.8  divide by (√(6^2 +5^2 ))=(√(36+25))=(√(61))  (6/(√(61)))cos 2a−(5/(√(61)))sin 2a=((1.8)/(√(61)))  let α such that sin α=(5/(√(61)))  cos αcos 2a−sin αsin 2a=((1.8)/(√(61)))  cos (α+2a)=((1.8)/(√(61)))  α+2a=2nπ±cos^(−1) ((1.8)/(√(61)))  0≤a≤180  a=π−(1/2)cos^(−1) ((1.8)/(√(61)))−(1/2)sin^(−1) (5/(√(61)))

$$\mathrm{6cos}\:\mathrm{2}{a}−\mathrm{5sin}\:\mathrm{2}{a}=\mathrm{1}.\mathrm{8} \\ $$$$\mathrm{divide}\:\mathrm{by}\:\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\sqrt{\mathrm{36}+\mathrm{25}}=\sqrt{\mathrm{61}} \\ $$$$\frac{\mathrm{6}}{\sqrt{\mathrm{61}}}\mathrm{cos}\:\mathrm{2}{a}−\frac{\mathrm{5}}{\sqrt{\mathrm{61}}}\mathrm{sin}\:\mathrm{2}{a}=\frac{\mathrm{1}.\mathrm{8}}{\sqrt{\mathrm{61}}} \\ $$$${let}\:\alpha\:{such}\:{that}\:\mathrm{sin}\:\alpha=\frac{\mathrm{5}}{\sqrt{\mathrm{61}}} \\ $$$$\mathrm{cos}\:\alpha\mathrm{cos}\:\mathrm{2}{a}−\mathrm{sin}\:\alpha\mathrm{sin}\:\mathrm{2}{a}=\frac{\mathrm{1}.\mathrm{8}}{\sqrt{\mathrm{61}}} \\ $$$$\mathrm{cos}\:\left(\alpha+\mathrm{2}{a}\right)=\frac{\mathrm{1}.\mathrm{8}}{\sqrt{\mathrm{61}}} \\ $$$$\alpha+\mathrm{2}{a}=\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}.\mathrm{8}}{\sqrt{\mathrm{61}}} \\ $$$$\mathrm{0}\leqslant{a}\leqslant\mathrm{180} \\ $$$${a}=\pi−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}.\mathrm{8}}{\sqrt{\mathrm{61}}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\sqrt{\mathrm{61}}} \\ $$

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