Question #2d3ed

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Answer 1 · 2018-02-13T03:13:48

$x = 0, \pm \frac{k π}{2}$ $x=0, +-(kpi)/2$

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$sec x = tan x + 1$ $secx=tanx+1$

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

Let's multiply both sides by $cos x$ $cosx$ :

$1 = sin x + cos x$ $1=sinx+cosx$

Let's raise both sides to the power of $2$ $2$ :

$1 = {(sin x + cos x)}^{2}$ $1=(sinx+cosx)^2$

${sin}^{2} x + {cos}^{2} x + 2 sin x cos x = 1$ $sin^2x+cos^2x+2sinxcosx=1$

But ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

$1 + 2 sin x cos x = 1$ $1+2sinxcosx=1$

$2 sin x cos x = 0$ $2sinxcosx=0$

We have a double-angle identity that says:

$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$

Therefore:

$sin 2 x = 0$ $sin2x=0$

$2 x = 0, \pm k π$ $2x=0, +-kpi$

$x = 0, \pm \frac{k π}{2}$ $x=0, +-(kpi)/2$