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veneciasp

veneciasp

Answered question

2022-07-04

Solve the equation 1 sin 2 k ( x ) + 1 cos 2 k ( x ) = 8 where k is an integer and x is a real number.

Answer & Explanation

poquetahr

poquetahr

Beginner2022-07-05Added 18 answers

We have
8 = 1 sin 2 k ( x ) + 1 cos 2 k ( x ) 2 | sin k ( x ) cos k ( x ) | = 2 k + 1 | sin k ( 2 x ) | 2 k + 1
Hence, we have k + 1 3 k 2
If k=0, we have 1 sin 2 k ( x ) + 1 cos 2 k ( x ) = 2
If k<0, we have
sin 2 k ( x ) + cos 2 k ( x ) sin 2 ( x ) + cos 2 ( x ) = 1
Hence, the only options are k=1 and k=2.
If k=2, we see that
sin 2 k ( x ) + cos 2 k ( x ) sin 2 ( x ) + cos 2 ( x ) = 1
This implies that
2 x = n π + π 2 x = n π 2 + π 4
When k=1, we have that
1 sin 2 ( x ) + 1 cos 2 ( x ) = 1 sin 2 ( x ) cos 2 ( x ) = 4 sin 2 ( 2 x ) = 8
Hence,
sin 2 ( 2 x ) = 1 2 2 x = n π ± π 4 x = n π 2 ± π 8

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