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lapo4ka [179]
2 years ago
5

9sin(2x) sin (x) = 9cos(x)

Mathematics
1 answer:
nikdorinn [45]2 years ago
8 0

Answer:

\sin(2x)  = 2 \sin(x)  \cos(x)  \\ 9( 2 \sin(x)  \cos(x) ) \sin(x)  = 9 \cos(x)  \\ 18 { \sin}^{2} (x)9 \cos(x)  - 9 \cos(x)  = 0 \\ 9 \cos(x) (2{ \sin}^{2} (x) - 1) = 0 \\ 9 \cos(x) = 0 \: or \: 2{ \sin}^{2} (x) - 1 = 0 \\  \sin(x)  =  \pm \frac{1}{ \sqrt{2} }  \\  \cos(x)  = 0 \\ x =  { \cos}^{ - 1} (0) \\ x =  \frac{ \pi}{2}  = 90 \degree \\ x = \frac{ \pi}{2} + 2\pi \: n \:  \forall \: n \:  \in \:  \Z  \\ = 90 \degree +  360\degree\: n \:  \forall \: n \:  \in \:  \Z \\  =  \pm\frac{ \pi}{4}  \pm2\pi \: n \:  \forall \: n \:  \in \:  \Z \\ x =  \pm45\degree  \pm 360\degree\: n \:  \forall \: n \:  \in \:  \Z

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Question 1)

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Question 2)

We have:

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Distribute:

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Distribute:

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