Show that the equation 1+sinxtanx=5cosx can be expressed as 6cos^2-cosx-1=0?

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Answer 1 · 2017-11-12T18:48:25

See the proof below

We need

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

${sin}^{2} x - {cos}^{2} x = 1$ $sin^2x-cos^2x=1$

Therefore,

$1 + sin x tan x = 5 cos x$ $1+sinxtanx=5cosx$

$1 + sin x \cdot \frac{sin x}{cos x} = 5 cos x$ $1+sinx*sinx/cosx=5cosx$

Multiply by $cos x$ $cosx$

$cos x + {sin}^{2} x = 5 {cos}^{2} x$ $cosx+sin^2x=5cos^2x$

Replacing ${sin}^{2} x$ $sin^2x$ by $1 - {cos}^{2} x$ $1-cos^2x$

$cos x + 1 - {cos}^{2} x = 5 {cos}^{2} x$ $cosx+1-cos^2x=5cos^2x$

$5 {cos}^{2} x - {cos}^{2} x - cos x - 1 = 0$ $5cos^2x-cos^2x-cosx-1=0$

$6 {cos}^{2} x - cos x - 1 = 0$ $6cos^2x-cosx-1=0$

$Q E D$ $QED$