How to solve 3sinx + 3cosx = secx + cosecx for x? Thank you!

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How to solve 3sinx + 3cosx = secx + cosecx for x? Thank you!

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Answer 1 · 2018-02-18T08:48:15

See the answer below...

Explanation:

$\Rightarrow 3 sin x + 3 cos x = sec x + csc x$ $=>3sinx+3cosx=secx+cscx$

$\Rightarrow 3 sin x + 3 cos x = \frac{1}{cos x} + \frac{1}{sin x}$ $=>3sinx+3cosx=1/cosx+1/sinx$

$\Rightarrow 3 sin x + 3 cos x = \frac{sin x + cos x}{sin x \cdot cos x}$ $=>3sinx+3cosx=(sinx+cosx)/(sinx cdot cosx$

$\Rightarrow (sin x + cos x) (3 - \frac{1}{sin x \cdot cos x}) = 0$ $=>(sinx+cosx)(3-1/(sinx cdot cosx))=0$

$\Rightarrow (sin x + cos x) (3 - \frac{2}{2 \cdot sin x \cdot cos x}) = 0$ $=>(sinx+cosx)(3-2/(2 cdot sinx cdot cosx))=0$

$\Rightarrow (sin x + cos x) (3 - \frac{2}{sin (2 x)}) = 0$ $=>(sinx+cosx)(3-2/sin(2x))=0$

Either,

$(sin x + cos x) = 0$ $(sinx+cosx)=0$

$\Rightarrow sin x = - cos x$ $=>sinx=-cosx$

$\Rightarrow tan x = - 1 = tan (- \frac{π}{4})$ $=>tanx=-1=tan(-pi/4)$

$color(red)(ul(bar(|color(green)(=>x=npi-pi/4)|$

Or,

$(3-2/sin(2x))=0$

$=>2/sin(2x)=3$

$=>sin2x=2/3=sinalpha$

$=>2x=npi+(-1)^nalpha$

$=>color(red)(ul(bar(|color(green)(x=(npi)/2+(-1)^n(alpha/2))|$

Hope it helps...
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How to solve 3sinx + 3cosx = secx + cosecx for x? Thank you!

1 Answer

Explanation:

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