Can somebody please solve question 11 for me?

Mathematics

1 answer:

love history [14]3 years ago

6 0

1+tan^2(A) = sec^2(A) [Pythagorean Identities]

tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)

*see photo for complete solution*

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Solve sin2∅=sin∅ on the interval 0≤x< 2<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align=

kondaur [170]

Answer:

$\large\boxed{b.\ 0,\ \pi,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}}$

Step-by-step explanation:

$\text{Use}\ \sin2x=2\sin x\cos x\\\\\sin2\O=\sin\O\\\\2\sin\O\cos\O=\sin\O\qquad\text{subtract}\ \sin\O\ \text{from both sides}\\\\2\sin\O\cos\O-\sin\O=0\qquad\text{distribute}\\\\\sin\O(2\cos\O-1)=0\iff\sin\O=0\ \vee\ 2\cos\O-1=0$

$\sin\O=0\iff\O=0\ \vee\ \O=\pi\\\\2\cos\O-1=0\qquad\text{add 1 to both sides}\\\\2\cos\O=1\qquad\text{divide both sides by 2}\\\\\cos\O=\dfrac{1}{2}\iff\O=\dfrac{\pi}{3}\ \vee\ \O=\dfrac{5\pi}{3}$

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${x}^{2} = 9 \\ \\ x = \sqrt{9} \\ \\ x = \sqrt{3 \times 3} \\ \\ x = 3$