Solving trigonometric equations

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SUMMARY

The discussion focuses on solving trigonometric equations, specifically (sin(3x))^2 + (cos(3x))^2 and 5sin(3x)^2 + 6cos(3x)^2. It is established that (sin(3x))^2 + (cos(3x))^2 equals one by applying the identity sin²θ + cos²θ = 1 with the substitution 3x = θ. For the expression 5sin(3x)^2 + 6cos(3x)^2, it can be simplified to 5 + cos²(3x) through algebraic manipulation.

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  • Understanding of trigonometric identities, specifically sin²θ + cos²θ = 1
  • Basic algebraic manipulation skills
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  • Knowledge of sine and cosine functions and their properties
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  • Learn about the unit circle and its role in trigonometric functions
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teng125
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as we know that (sin(x))^2 + (cos(x))^2=1
how about (sin(3x))^2 + (cos(3x))^2 and 5sin(3x)^2 + 6cos(3x)^2??

how can we solve these problems??
thanx
 
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What are you trying to solve? You could substitute numbers for the x's and compute the value.

I don't believe there's a nifty identity for (sin(3x))^2 + (cos(3x))^2 even though it does look somewhat similar to (sin(x))^2 + (cos(x))^2.
 
Actually, aslong as the arguments match it equals one. So (sin(3x))^2 + (cos(3x))^2 does equal one.
 
Indeed, if we apply the identity \sin^{2}\theta +\cos^{2}\theta = 1,to the above expression and simply use the substitution 3x = \theta... :wink:
 
For 5sin(3x)^2 + 6cos(3x)^2, write it as 5sin^3(3x)+ 5cos^2(3x)+ cos^2(3x)= 5(sin^3(3x)+ cos^2(3x))+ cos^2(3x)= 5+ cos^2(3x).
 

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