Solve the trigonometric equation sin x - 2 cos 2x = 1 and give the general solution.

Step-by-step solution

1. 1

   Rewrite everything in terms of one trigonometric function. The equation mixes the angle $x$ with the angle $2x$, which cannot be solved directly. Of the three forms of the double-angle identity, choose the one written purely in sine, since the other term is $\sin x$:

   $\cos 2x=1-2\sin^{2}x.$

2. 2

   Substitute and collect.

   $\sin x-2\left(1-2\sin^{2}x\right)=1\ \Longrightarrow\ \sin x-2+4\sin^{2}x=1,$

   $4\sin^{2}x+\sin x-3=0.$

   The equation is now a quadratic in a single quantity.

3. 3

   Solve the quadratic. Put $t=\sin x$, so $4t^{2}+t-3=0$. The discriminant is $D=1^{2}-4(4)(-3)=49$, a perfect square, so

   $t=\frac{-1\pm 7}{8}\ \Longrightarrow\ t_{1}=\frac{3}{4},\qquad t_{2}=-1.$

   Both values lie in $[-1,1]$, so neither has to be discarded — always check this, because a root outside $[-1,1]$ would be extraneous.

4. 4

   Solve $\sin x=\tfrac34$. This value is not a standard angle, so the answer stays in terms of $\arcsin$. Sine takes the value $\tfrac34$ once in the first quadrant and once in the second:

   $x=\arcsin\frac{3}{4}+2\pi n\quad\text{or}\quad x=\pi-\arcsin\frac{3}{4}+2\pi n,\qquad n\in\mathbb{Z},$

   which the compact formula $x=(-1)^{n}\arcsin\frac{3}{4}+\pi n$ packs into one line. Numerically $\arcsin\tfrac34\approx 0.8481$ rad $\approx 48.59^{\circ}$.

5. 5

   Solve $\sin x=-1$. Here the sine hits its minimum, which happens at a single point per period, so there is only one family:

   $x=-\frac{\pi}{2}+2\pi n\quad\left(\text{equivalently }x=\frac{3\pi}{2}+2\pi n\right),\qquad n\in\mathbb{Z}.$

6. 6

   Check both families in the original equation. At $x=\arcsin\tfrac34\approx 0.8481$: $\sin x-2\cos 2x=0.75-2(-0.125)=1$. At $x=-\tfrac{\pi}{2}$: $-1-2\cos(-\pi)=-1+2=1$. Both satisfy the equation, so the solution set is the union of the two families.