How do I solve equations like 3sin^2(x) - 2cos(x) = 2

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To solve equations like this we need to get the equation in terms of one trigonometric function, ie one of sin(x), cos(x) or tan(x)

It is usually easiest to get everything in terms of whatever the linear part of the equation is, in this case cos(x).

To do this we use that sin2(x) + cos2(x) = 1. Since we're trying to get rid of the sin2(x) part of the equation we rearrange this to get that sin2(x) = 1 - cos2(x)

We substitute this into the equation 3sin2(x) - 2cos(x) = 2

This gives 3(1 - cos2(x)) - 2cos(x) = 2

This then becomes 3 - 3cos2(x) - 2cos(x) = 2

which we can rearrange to 3cos2(x) + 2cos(x) - 1 = 0

Now that we have an equation just in terms of one trigonometric function , here cos(x), we can just replace it with some letter, say y

Thus this is just 3y2 + 2y - 1 = 0. We can solve this with the usual quadratic equation, giving y = -1 and y = 1/3. As y = cos(x) this is just cos(x) = -1 and cos(x) = 1/3. To find x we then just solve these in the usual way

Elliot B. A Level Maths tutor, A Level Further Mathematics  tutor, GC...

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