Question

How do you find the general solutions for `sinx = cos2x`?

Answer 1

`x=30^{\circ}, 270^{\circ}` `[0^0 \leq x \leq \360^0]`

Explanation:

We know,

`cos2x=cos^2x-sin^2x=1-2sin^2x`

So, let's solve the equation now,

`sinx=cos2x=1-2sin^2x`
`\rightarrow 2sin^2x+sinx-1=0`
`\rightarrow 2sin^2x+2sinx-sinx-1=0`
`\rightarrow 2sinx(sinx+1)-1(sinx+1)=0`
`\rightarrow (2sinx-1)(sinx+1)=0`

Now,

`2sinx-1=0`
`\rightarrow sinx=\frac{1}{2}`
`\rightarrow x=sin^{-1}(` `\frac{1}{2}`)
`\rightarrow x=30^{\circ}`

And, `sinx+1=0`
`\rightarrow x=sin^{-1}(-1)` `= 270^{\circ}`

As we just need the general solutions, we should take only this two values as the general solutions .
Answer : `30^0, 270^0`
That's it!

Answer 2

Solve sin x = cos 2x

Explanation:

Apply trig identity: `cos 2x = 1 - 2sin^2 x`
`sin x = 1 - 2sin^2 x`. Solve the quadratic equation:
`2sin^2 x + sin x - 1 = 0`
Since (a - b + c = 0), use Shortcut. Two real roots: sin x = -1 and `sin x = -c/a = 1/2`.

`sin x = 1/2` --> x = 30 deg and x = 150 deg `(pi/6 and (5pi)/6)`
sin x = -1 --> x = 270 deg `((3pi)/2)`
General solutions:
x = 30 + k360 deg
x = 150 + k360
x = 270 + k360.