Need help solving this trig equation

AI Thread Summary
To solve the equation 3cos(θ) + 1.595*sin(θ) = 3.114, the identity (sinθ)^2 + (cosθ)^2 = 1 is applied, leading to an incorrect result of cosθ = +/- 1.0526, which is outside the valid range for cosine. A suggested alternative method involves rewriting the equation in the form a cos(θ) + b sin(θ) = R cos(θ - φ), where R = √(a² + b²) and φ = arctan(b/a). This transformation simplifies the problem and helps find the correct angle θ. The discussion emphasizes the importance of ensuring results stay within the defined ranges for trigonometric functions. Understanding these identities is crucial for solving trigonometric equations effectively.
flgdx
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Homework Statement


what's the best way to solve this equation: 3cos(θ) + 1.595*sin(θ) = 3.114

Homework Equations


(sinθ)^2 + (cosθ)^2 = 1

The Attempt at a Solution


I tried using the identity above to solve this equation and ended up with cosθ = +/- 1.0526.
 
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Could you show your steps for your solution?
 
Comeback City said:
Could you show your steps for your solution?

so since (sinθ)^2 + (cosθ)^2 = 1, I solved for sinθ and got sqrt(1 - (cosθ)^2). Then I subtituted it into the equation 3cos(θ) + 1.595*sin(θ) = 3.114 and solved for θ.
 
flgdx said:

Homework Statement


what's the best way to solve this equation: 3cos(θ) + 1.595*sin(θ) = 3.114

Homework Equations


(sinθ)^2 + (cosθ)^2 = 1

The Attempt at a Solution


I tried using the identity above to solve this equation and ended up with cosθ = +/- 1.0526.
Well, that's obviously wrong because -1 ≤ cos θ ≤ 1, and 1.0526 is outside that range. Your approach should work though.

Another approach is to use
$$a \cos\theta + b \sin\theta = \sqrt{a^2+b^2} \cos(\theta-\phi)$$ where ##\tan\phi = b/a##.
 
could you pls show me how you got that equation?
 
Consider a right triangle below and use the identity ##\cos(\theta-\phi) = \cos\theta\,\cos\phi + \sin\theta\,\sin\phi##.

triangle.png
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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