How do you solve 2sin^2x+7sinx=4 in the interval [0,360]?

1 Answer
Alan P.
Dec 14, 2016

x in {30^@, 150^@, 194.48^@,345.52^@)

Explanation:

Given
color(white)("XXX")2sin^2(x)+7sin(x)=4
Rearranging into standard quadratic form:
color(white)("XXX")2sin^2(x)+7sin(x)-4=0
Factoriing:
color(white)("XXX")(2sin(x)-1)(sin(x)+4)=0
rarr
color(white)("XXX"){: (sin(x)=1/2," or ",sin(x)=-1/4), ("in the range",[0,360^@],), (x=30^@" or "150^@,,x~~194.48^@" or "345.52^@) :}

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Note: x is one of the standard angles for sin(x)=1/2;
but I needed to use a calculator for arcsin(-1/4) to evaluate sin(x)=-1/4

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