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Answer 1
First, we rewrite the given equation: $ \cos(2\theta) + \sin(3\theta) = 1 $
We know that \( \cos(2\theta) = \cos^2(\theta) – \sin^2(\theta) \) and \( \sin(3\theta) = 3\sin(\theta) – 4\sin^3(\theta) \).
Combining these identities: $ \cos^2(\theta) – \sin^2(\theta) + 3\sin(\theta) – 4\sin^3(\theta) = 1 $
This equation is complex and needs to be solved numerically. Let’s solve for specific values:
Approximating using numerical methods, we find: $ \theta \approx 0.4516, 2.6902, 4.8381 $
Answer 2
First approach the given trigonometric equation: $ cos(2 heta) + sin(3 heta) = 1 $
We can transform the equation into terms of ( sin ) and ( cos ) of single angle using double angle and triple angle formulas:
The double angle formula is ( cos(2 heta) = 2cos^2( heta) – 1 ), and the triple angle formula is ( sin(3 heta) = 3sin( heta) – 4sin^3( heta) ).
Thus, the equation becomes: $ 2cos^2( heta) – 1 + 3sin( heta) – 4sin^3( heta) = 1 $
By substituting ( cos heta = sqrt{1-sin^2 heta} ), we solve:
Numerically finding solutions: $ heta approx 0.4516, 2.6902, 4.8381 $
Answer 3
Given equation: $ cos(2 heta) + sin(3 heta) = 1 $
Using double and triple angle formulas:
$ cos(2 heta) = cos^2( heta) – sin^2( heta) $
$ sin(3 heta) = 3sin( heta) – 4sin^3( heta) $
Thus: $ cos^2( heta) – sin^2( heta) + 3sin( heta) – 4sin^3( heta) = 1 $
Solutions: $ heta approx 0.4516, 2.6902, 4.8381 $
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