请求协助求解三角函数方程：a sin(x+θ)+b cos(2x)=c
Let's walk through how to solve this trigonometric equation step by step—no need to overcomplicate things. The equation we're tackling is:
$$a\sin(x+\theta)+b\cos(2x)=c$$
where $a, b, c, \theta$ are all known constants. Here's a structured approach to find all real solutions for $x$:
Step 1: Expand the sine of a sum
First, use the sine addition identity to break down $\sin(x+\theta)$. Remember that:
$$\sin(A+B) = \sin A \cos B + \cos A \sin B$$
Applying this gives:
$$\sin(x+\theta) = \sin x \cos \theta + \cos x \sin \theta$$
Substitute this back into the original equation:
$$a\left(\sin x \cos \theta + \cos x \sin \theta\right) + b\cos(2x) = c$$
Step 2: Rewrite $\cos(2x)$ using a double-angle identity
We have a few options for $\cos(2x)$, but the most useful here is converting it to a function of $\sin x$ (or $\cos x$) to set up a solvable polynomial. Let's use:
$$\cos(2x) = 1 - 2\sin^2 x$$
Substituting this in gives:
$$a\cos\theta \cdot \sin x + a\sin\theta \cdot \cos x + b\left(1 - 2\sin^2 x\right) = c$$
Rearrange terms to group like terms together:
$$-2b\sin^2 x + a\cos\theta \cdot \sin x + a\sin\theta \cdot \cos x + (b - c) = 0$$
Step 3: Eliminate the $\cos x$ term (with a caveat)
The problem now is we have both $\sin x$ and $\cos x$ terms. To fix this, we can square both sides of the equation after isolating the linear trigonometric terms. First, rearrange the equation to get the $\sin x$ and $\cos x$ terms on one side:
$$a\cos\theta \cdot \sin x + a\sin\theta \cdot \cos x = c - b + 2b\sin^2 x$$
Let $t = \sin x$, so $\cos x = \pm\sqrt{1 - t^2}$. Substitute these in:
$$a\cos\theta \cdot t + a\sin\theta \cdot \left(\pm\sqrt{1 - t^2}\right) = c - b + 2bt^2$$
Now isolate the square root term and square both sides to eliminate it:
$$a\sin\theta \cdot \left(\pm\sqrt{1 - t^2}\right) = (c - b + 2bt^2) - a\cos\theta \cdot t$$
$$a2\sin2\theta \cdot (1 - t^2) = \left[(c - b + 2bt^2) - a\cos\theta \cdot t\right]^2$$
Step 4: Solve the resulting quartic equation
Expanding both sides of the equation will give you a quartic (4th-degree) polynomial in $t$. Let's expand the right-hand side first:
$$(c - b + 2bt^2 - a\cos\theta \cdot t)^2 = (2bt^2 - a\cos\theta \cdot t + (c - b))^2$$
When you expand this and combine like terms with the left-hand side, you'll end up with an equation of the form:
$$At^4 + Bt^3 + Ct^2 + Dt + E = 0$$
where $A, B, C, D, E$ are constants derived from $a, b, c, \theta$.
You can solve this quartic using algebraic methods (like Ferrari's method) or, for practical purposes, use numerical root-finding tools (since quartic solutions can get messy with arbitrary constants).
Important note:
Only keep real roots $t$ where $|t| \leq 1$—because $\sin x$ can never be outside the range $[-1, 1]$.
Step 5: Convert back to $x$ and check for extraneous solutions
For each valid root $t_i$:
The general solutions for $x$ are:
$$x = \arcsin(t_i) + 2k\pi \quad \text{or} \quad x = \pi - \arcsin(t_i) + 2k\pi$$
where $k$ is any integer (this accounts for all periodic solutions).Critical: Since we squared the equation earlier, some of these solutions might not satisfy the original equation. Plug each candidate $x$ back into the original equation to verify it's a valid solution.
Alternative: Numerical methods for faster solutions
If you don't need an algebraic closed-form solution, numerical methods like the Newton-Raphson method are often easier to implement. Define the function:
$$f(x) = a\sin(x+\theta) + b\cos(2x) - c$$
Pick initial guesses for $x$ (you can plot $f(x)$ to find approximate root locations) and iterate using:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
where $f'(x) = a\cos(x+\theta) - 2b\sin(2x)$ is the derivative of $f(x)$. Continue iterating until the value of $f(x)$ is close enough to 0 (within your desired tolerance).
内容的提问来源于stack exchange，提问作者Mohsen djalali khalilabadi
