Solve the equation ${tan}^{2} x = 3 (3 - sec x)$ $tan^2x=3(3-secx)$ , if $x$ $x$ lies in interval $[0^{\circ}, 360^{\circ})$ $[0^@,360^@)$ ?

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Answer 1 · 2017-02-05T14:43:57

$x$ $x$ is $60^{\circ}, {101.5}^{\circ}, {258.5}^{\circ}$ $60^@, 101.5^@, 258.5^@$ or $300^{\circ}$ $300^@$

As ${tan}^{2} x = 3 (3 - sec x)$ $tan^2x=3(3-secx)$ , we have

${sec}^{2} x - 1 = 9 - 3 sec x$ $sec^2x-1=9-3secx$

or ${sec}^{2} x + 3 sec x - 10 = 0$ $sec^2x+3secx-10=0$

or $(sec x + 5) (sec x - 2) = 0$ $(secx+5)(secx-2)=0$

Hence $sec x = - 5$ $secx=-5$ or $sec x = 2$ $secx=2$ and using tables

$cos x = - 0.2 = {cos 101.5}^{\circ}$ $cosx=-0.2=cos101.5^@$ and ${cos 258.5}^{\circ}$ $cos258.5^@$

or $cos x = 0.5 = {cos 60}^{\circ}$ $cosx=0.5=cos60^@$ and ${cos 300}^{\circ}$ $cos300^@$

(as $cos (360^{\circ} - A) = cos A$ $cos(360^@-A)=cosA$ )

Hence, $x$ $x$ is $60^{\circ}, {101.5}^{\circ}, {258.5}^{\circ}$ $60^@, 101.5^@, 258.5^@$ or $300^{\circ}$ $300^@$

Solve the equation tan^2x=3(3-secx)tan2x=3(3−secx)tan^2x=3(3-secx), if xxx lies in interval [0^@,360^@)[0∘,360∘)[0^@,360^@)?