Geometry and Trigonometry - More trigonometric equations – identities

Solve the equation $\cos 2x = \sin x$ for $0 \le x \le 2\pi$.

$1 - 2\sin^2 x = \sin x$ $2\sin^2 x + \sin x - 1 = 0$ $(2\sin x - 1)(\sin x + 1) = 0$ $\sin x = \frac{1}{2} \text{ or } \sin x = -1$ For $\sin x = \frac{1}{2}$, $x = \frac{\pi}{6}, \frac{5\pi}{6}$. For $\sin x = -1$, $x = \frac{3\pi}{2}$. The solution set is $x \in \{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\}$.

First, the double angle identity $\cos 2x = 1 - 2\sin^2 x$ is used to express the entire equation in terms of $\sin x$. The resulting quadratic equation is factored. Then, the basic trigonometric equations are solved within the given domain.

Solve $2\cos^2 x + 3\sin x - 3 = 0$ for $0^\circ \le x \le 360^\circ$.

$2(1 - \sin^2 x) + 3\sin x - 3 = 0$ $2 - 2\sin^2 x + 3\sin x - 3 = 0$ $-2\sin^2 x + 3\sin x - 1 = 0$ $2\sin^2 x - 3\sin x + 1 = 0$ $(2\sin x - 1)(\sin x - 1) = 0$ $\sin x = \frac{1}{2} \implies x = 30^\circ, 150^\circ$ $\sin x = 1 \implies x = 90^\circ$

The Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ is substituted into the equation to create a quadratic in terms of $\sin x$. After simplifying and factoring, the values for $x$ are found in degrees.

Given that $\sin A = \frac{3}{5}$ and $A$ is obtuse, find the exact value of $\sin 2A$.

Since $\sin^2 A + \cos^2 A = 1$: $\cos^2 A = 1 - (\frac{3}{5})^2 = \frac{16}{25}$ Since $A$ is obtuse (Quadrant II), $\cos A$ must be negative: $\cos A = -\frac{4}{5}$ Now use the double angle formula: $\sin 2A = 2\sin A \cos A = 2(\frac{3}{5})(-\frac{4}{5}) = -\frac{24}{25}$

To find $\sin 2A$, we need both $\sin A$ and $\cos A$. We find $\cos A$ using the Pythagorean identity, being careful to choose the negative root because the angle is in the second quadrant. Finally, we apply the double angle identity.

1. Use the double angle identity $\sin 2x = 2 \sin x \cos x$ and the identity $\tan x = \frac{\sin x}{\cos x}$: $2 \sin x \cos x = \frac{\sin x}{\cos x}$
2. Rearrange the equation (multiply by $\cos x$, noting $\cos x \neq 0$): $2 \sin x \cos^2 x - \sin x = 0$
3. Factor out $\sin x$: $\sin x (2 \cos^2 x - 1) = 0$
4. Set each factor to zero: $\sin x = 0 \implies x = 0, \pi$ $2 \cos^2 x - 1 = 0 \implies \cos^2 x = \frac{1}{2} \implies \cos x = \pm \frac{1}{\sqrt{2}}$
5. Solve for $x$ in the interval $[0, \pi]$: For $\cos x = \frac{1}{\sqrt{2}}$, $x = \frac{\pi}{4}$ For $\cos x = -\frac{1}{\sqrt{2}}$, $x = \frac{3\pi}{4}$ Final solutions: $x \in \{0, \frac{\pi}{4}, \frac{3\pi}{4}, \pi\}$.

This example uses the substitution of both a double angle identity and the tangent identity to factorize the equation. It is important to factorize rather than divide by $\sin x$ to avoid losing solutions.

1. Replace $\cos 2x$ with the form that involves only $\cos x$: $3(2\cos^2 x - 1) - \cos x + 1 = 0$
2. Expand and simplify to form a quadratic equation: $6\cos^2 x - 3 - \cos x + 1 = 0$ $6\cos^2 x - \cos x - 2 = 0$
3. Let $u = \cos x$. Factor the quadratic $6u^2 - u - 2 = 0$: $(3u - 2)(2u + 1) = 0$
4. Solve for $u$: $u = \frac{2}{3}$ or $u = -\frac{1}{2}$
5. Find values of $x$ for $\cos x = \frac{2}{3}$: $x \approx 0.841, 5.44$ (using calculator)
6. Find values of $x$ for $\cos x = -\frac{1}{2}$: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$ ($x \approx 2.09, 4.19$) Final solution set: $x \approx \{0.841, 2.09, 4.19, 5.44\}$.

By choosing the $\cos 2x = 2\cos^2 x - 1$ identity, we transform the equation into a standard quadratic form which can be factored.