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2sin2x-cosx=0
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Do you want the answer or how to do it?
how to solve it and how to do it
In the end you want the equation to be A - A= 0 where A is any number. So what value of x will give you the same answer for both the 2sin2x and cos x?
\(\bf {\color{brown}{ sin(2\theta)}}=2sin(\theta)cos(\theta) \\ \quad \\ \quad \\ 2sin(2x)-cos(x)=0\implies 2\cdot {\color{brown}{ 2sin(x)cos(x)}}=cos(x) \\ \quad \\ 4sin(x)=\cancel{ \cfrac{cos(x)}{cos(x)} }\) solve for sin(x), then take arcsine to both sides to get "x"
3 sin^2x-7 sinx+2=0 ?
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?
3 sin(squared)x- 7sin x=2=0
well, "x" is not that
cos x sin x- sinx=0 solve the equation in the interval [0,2pi)
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