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Question
how i can solve this equation?
Practically (how i can get theta1 and theta2 in terms of theta?)
3cos(theta1)+4cos(theta2)=3-2cos(theta)
3sin(theta1)+4sin(theta2)=1.95-2sin(theta)
Many thanks
By this do you mean
3cosA + 4cosB = 3 - 2cosC
3sinA + 4sinB = 1.95 - 2sinC
where as
A = theta1
B = theta2
C = theta
of which A, B, and C are not neccessarily equal.
or do you mean
3cosA + 4cosA = 3 - 2cosA
3sinA + 4sinA = 1.95 - 2sinA
whereas A stands for Theta.
if you can answer that, then i should be able to help you out.
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Answers by Expert:
Sherman D.
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I can answer questions dealing in mathematics of all kinds except for Physics and Calculus, but i can answer questions in Pre-Calculus and Chemistry. I can also answer questions in Recipes of all kinds. I can find games cheats/walkthroughs, but i can`t find a specific game online or offline. I can also do history and recipes for alcoholic beverages.
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