Soal 40

Jika $\sin 2x + \cos 2x = -16 \cos x + 8 \sin x + \cos^2 x$ dengan $0 \leq x \leq \frac{\pi}{2}$ , maka $\sin 2x$ = ....

Gunakan identitas trigonometri berikut

\sin 2x = 2 \sin x \cos x

\cos 2x = 2 \cos^2 x - 1

Maka

\sin 2x + \cos 2x = -16 \cos x + 8 \sin x + \cos^2 x

2 \sin x \cos x + 2 \cos^2 x - 1 = -16 \cos x + 8 \sin x + \cos^2 x

2 \sin x \cos x + 2 \cos^2 x - 1 + 16 \cos x - 8 \sin x - \cos^2 x = 0

2 \sin x \cos x + \cos^2 x + 16 \cos x - 8 \sin x - 1 = 0

2 \cos x (\sin x + 8) - \sin^2 x - 8 \sin x = 0

2 \cos x (\sin x + 8) - \sin x (\sin x + 8) = 0

(2 \cos x - \sin x)(\sin x + 8) = 0

Maka $2 \cos x - \sin x = 0$ atau $\sin x + 8 = 0$ .

Karena $0 \leq x \leq \frac{\pi}{2}$ , maka $\sin x \geq 0$ , sehingga $\sin x + 8 \neq 0$ .

Jadi $2 \cos x - \sin x = 0$

\sin x = 2 \cos x

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

\tan x = 2 = \frac{2}{1}

Sisi depannya adalah $2$ , dan sisi sampingnya $1$ . Maka sisi miringnya adalah

\text{Mi} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}

Dengan begitu

\sin x = \frac{\text{de}}{\text{mi}} = \frac{2}{\sqrt{5}}

\cos x = \frac{\text{sa}}{\text{mi}} = \frac{1}{\sqrt{5}}

Maka nilai $\sin 2x$

\sin 2x = 2 \sin x \cos x

\sin 2x = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}

\sin 2x = \frac{4}{5}

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