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3cos^2x+sin^2x+5sinx=0

If sin^2x+cos^2x=1

Then wouldnt this be 3+5sinx=0?
Reply 1
Avatar for Smaug123
Smaug123
15
No, sorry - because 3cos⁡2(x)+sin⁡2(x)=cos⁡2(x)+sin⁡2(x)+2cos⁡2(x)=1+2cos⁡2(x)3 \cos^2(x) + \sin^2(x) = \cos^2(x)+\sin^2(x)+2 \cos^2(x) = 1+2 \cos^2(x).
Reply 2
Avatar for Tynos
Tynos
OP
2
Original post by Smaug123
No, sorry - because 3cos⁡2(x)+sin⁡2(x)=cos⁡2(x)+sin⁡2(x)+2cos⁡2(x)=1+2cos⁡2(x)3 \cos^2(x) + \sin^2(x) = \cos^2(x)+\sin^2(x)+2 \cos^2(x) = 1+2 \cos^2(x).


Ty.
Reply 3
Avatar for ztibor
ztibor
10
Original post by Tynos
3cos^2x+sin^2x+5sinx=0

If sin^2x+cos^2x=1

Then wouldnt this be 3+5sinx=0?


There is both sine and cosine function in the equation.
The solution will more simple if there is only sine or cosine functions in this equation.
From the identity you wrote
cos^2x=1-sin^2x
SUb this in cos^2x and you will get a quadratic equation for sinx
For sinx you can use the formula for quadratic equations.
sin⁡x1,2=−b±b2−4ac2a\sin x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
where a,b,c is the factor of proper sine-power or constant.
(edited 10 years ago)

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