12=sin2xtanxWhat would be my next step to get to&nbsp;x=45°?

Let t=tan⁡x. Since 12=tsin2x=t3cos2x=t31+t2 0=2t3−t2−1=(t−1)(2t2+t+1) Since 2t2+t+1&gt;0,t=1. This is achieved when x=(45+180n)∘ for n∈Z

sin2xtan⁡x=sin2xsin⁡xcos⁡x=sin⁡xcos⁡x sin⁡xcos⁡x=12sin⁡2x 12sin⁡2x=12⇔sin⁡2x=1⇔2x=90∘+k360∘ x=45∘+k180∘

cancel sin⁡x on either side, simplify sin⁡2x=1→2x=π2±2kπ→x=π4±kπ

Let t=tan⁡x. Since 12=tsin2x=t3cos2x=t31+t2 0=2t3−t2−1=(t−1)(2t2+t+1) Since 2t2+t+1&gt;0,t=1. This is achieved when x=(45+180n)∘ for n∈Z

sin2xtan⁡x=sin2xsin⁡xcos⁡x=sin⁡xcos⁡x sin⁡xcos⁡x=12sin⁡2x 12sin⁡2x=12⇔sin⁡2x=1⇔2x=90∘+k360∘ x=45∘+k180∘

cancel sin⁡x on either side, simplify sin⁡2x=1→2x=π2±2kπ→x=π4±kπ

Expert answers to "12=sin2xtanxWhat would be my next step to get..."

Donald Johnson

2022-01-02

$\frac{1}{2} = \frac{\sin^{2} x}{\tan x}$

What would be my next step to get to $x = 45 °$ ?

Tiefdruckot

Beginner2022-01-03Added 46 answers

Let

t = \tan x

. Since

\frac{12}{=} t \sin^{2} x = t^{3} \cos^{2} x = \frac{t^{3}}{1 + t^{2}}

0 = 2 t^{3} - t^{2} - 1 = (t - 1) (2 t^{2} + t + 1)

Since

2 t^{2} + t + 1 > 0, t = 1

. This is achieved when

x = {(45 + 180 n)}^{\circ}

for

n \in Z

puhnut1m

Beginner2022-01-04Added 33 answers

\frac{\sin^{2} x}{\tan x} = \frac{\sin^{2} x}{\frac{\sin x}{\cos x}} = \sin x \cos x

\sin x \cos x = \frac{12}{\sin 2} x

\frac{12}{\sin 2} x = \frac{12}{\Leftrightarrow} \sin 2 x = 1 \Leftrightarrow 2 x = 90^{\circ} + k 360^{\circ}

x = 45^{\circ} + k 180^{\circ}

Vasquez

Expert2022-01-09Added 669 answers

cancel $\sin x$ on either side, simplify
$\sin 2 x = 1 \to 2 x = \frac{π}{2} \pm 2 k π \to x = \frac{π}{4} \pm k π$

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