$6 {(tan x)}^{2} - 4 {(sin x)}^{2} = 1$ $6(tanx)^2-4(sinx)^2=1$ . Solve x?

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Answer 1 · 2017-10-11T15:14:47

$6 {(tan x)}^{2} - 4 {(sin x)}^{2} = 1$ $6(tanx)^2-4(sinx)^2=1$

$\Rightarrow 6 ({tan}^{2} x) \times {cos}^{2} x - 4 {sin}^{2} x \times {cos}^{2} x = {cos}^{2} x$ $=>6(tan^2x)xxcos^2x-4sin^2x xxcos^2x=cos^2x$

$\Rightarrow 6 {sin}^{2} x - 4 {sin}^{2} x \times {cos}^{2} x = {cos}^{2} x$ $=>6sin^2x-4sin^2x xxcos^2x=cos^2x$

$\Rightarrow 6 (1 - {cos}^{2} x) - 4 (1 - {cos}^{2} x) {cos}^{2} x = {cos}^{2} x$ $=>6(1-cos^2x)-4(1-cos^2x) cos^2x=cos^2x$

$\Rightarrow 6 - 6 {cos}^{2} x - 4 {cos}^{2} x + 4 {cos}^{4} x = {cos}^{2} x$ $=>6-6cos^2x-4cos^2x+4cos^4x=cos^2x$

$\Rightarrow 4 {cos}^{4} x - 11 {cos}^{2} x + 6 = 0$ $=>4cos^4x-11cos^2x+6=0$

$\Rightarrow 4 {cos}^{4} x - 8 {cos}^{2} x - 3 {cos}^{2} x + 6 = 0$ $=>4cos^4x-8cos^2x-3cos^2x+6=0$

$\Rightarrow 4 {cos}^{2} x ({cos}^{2} x - 2) - 3 ({cos}^{2} x - 2) = 0$ $=>4cos^2x(cos^2x-2)-3(cos^2x-2)=0$

$= ({cos}^{2} x - 2) (4 {cos}^{2} x - 3) = 0$ $=(cos^2x-2)(4cos^2x-3)=0$

when ${cos}^{2} x - 2 = 0 \Rightarrow cos x \pm \sqrt{2}$ $cos^2x-2=0=>cosxpmsqrt2$

but $- 1 \leq cos x \leq + 1$ $-1<=cosx <=+1$
so this solution is not acceptable.

when

$(4 {cos}^{2} x - 3) = 0$ $(4cos^2x-3)=0$

$cos x = \pm \frac{\sqrt{3}}{2}$ $cosx=pmsqrt3/2$

For

$cos x = + \frac{\sqrt{3}}{2} = cos (\frac{π}{6})$ $cosx=+sqrt3/2=cos(pi/6)$

$=>x=2npipmpi/6" where " n in ZZ$

For

$cosx=-sqrt3/2=-cos(pi/6)=cos((5pi)/6)$

$=>x=2npipm(5pi)/6" where " n in ZZ$

6(tanx)^2-4(sinx)^2=16(tanx)2−4(sinx)2=16(tanx)^2-4(sinx)^2=1. Solve x?