# How do you solve #cos2x+2sinx=0# using the double angle identity?

Solve f(x) = cos 2x + 2sin x = 0

Ans: 201.47 and 338.53 deg

Check: x = 201,47 --> 2x = 402.94 = 42.94 + 360 --> cos 2x = cos 42.94 = 0.73 --> 2sin 201.47 = -0.73. cos 2x - 2sin x = 0.73 - 0.73 = 0. OK

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To solve the equation ( \cos(2x) + 2\sin(x) = 0 ) using the double angle identity, we first rewrite ( \cos(2x) ) in terms of ( \cos(x) ) and ( \sin(x) ). The double angle identity for cosine is ( \cos(2x) = 2\cos^2(x) - 1 ).

Substitute this identity into the equation:

[ 2\cos^2(x) - 1 + 2\sin(x) = 0 ]

Next, replace ( \cos^2(x) ) with ( 1 - \sin^2(x) ) using the Pythagorean identity ( \cos^2(x) + \sin^2(x) = 1 ):

[ 2(1 - \sin^2(x)) - 1 + 2\sin(x) = 0 ]

Now, distribute and rearrange terms:

[ 2 - 2\sin^2(x) - 1 + 2\sin(x) = 0 ] [ -2\sin^2(x) + 2\sin(x) + 1 = 0 ]

Multiply the entire equation by -1 to make it easier to work with:

[ 2\sin^2(x) - 2\sin(x) - 1 = 0 ]

This quadratic equation can be solved using the quadratic formula:

[ \sin(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

In this equation, ( a = 2 ), ( b = -2 ), and ( c = -1 ). Substituting these values:

[ \sin(x) = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(-1)}}{2(2)} ] [ \sin(x) = \frac{2 \pm \sqrt{4 + 8}}{4} ] [ \sin(x) = \frac{2 \pm \sqrt{12}}{4} ] [ \sin(x) = \frac{2 \pm 2\sqrt{3}}{4} ] [ \sin(x) = \frac{1 \pm \sqrt{3}}{2} ]

The solutions for ( \sin(x) ) are ( \frac{1 + \sqrt{3}}{2} ) and ( \frac{1 - \sqrt{3}}{2} ). However, we need to check if these values satisfy the original equation ( \cos(2x) + 2\sin(x) = 0 ).

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