Trigonometric Functionssimplify sin squared functions

AI Thread Summary
The discussion focuses on simplifying the trigonometric equation 2sin^2x - 3sinx - 2 = 0. It is suggested to treat the equation as a quadratic by substituting sinx with x, resulting in 2x^2 - 3x - 2 = 0. The equation can be factored into (2x + 1)(x - 2) and then substituting back gives (2sinx + 1)(sinx - 2) = 0. The conversation also notes that if using this method is not allowed, further assistance may be needed for applying trigonometric identities. Overall, the simplification approach is validated through standard quadratic methods.
ku1005
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just looking at another question to do with trigonometric functions and I can't see how they simplify the follwing:
2sin^2x-3sinx-2=0 to
(2sinx+1)(sinx-2)=0
again i prob thinking sumthin really stupid...but i can't see wat! cheers
 
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The equation can be treated like a normal quadratic equation.

use substitution and replace "sinx" with "x"
2(sinx)^2-3sinx-2=0 ----> 2x^2-3x-2=0

factor the equation
2x^2-3x-2=0 = (2x+1)(x-2)

Replace "x" with "sinx"
(2x+1)(x-2)=0 -----> (2sinx+1)(sinx-2)=0


If you aren't allowed to use this method and you have to use trig identities then someone else will have to help. I'm too lazy to look up or derive the identities right now :-P
 
cheers mate...and i like ur thinkin! LOL
 

Thread 'Greatest possible value of a constant in polynomial'

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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