Math Problem Statement

Solve for x: 3Cos x + Sin x = 1

Solution

To solve the equation 3cosx+sinx=13\cos x + \sin x = 1, we can proceed by expressing the equation in a single trigonometric form.


Steps to Solve 3cosx+sinx=13\cos x + \sin x = 1

  1. Combine 3cosx3\cos x and sinx\sin x into one term: We use the concept of expressing acosx+bsinxa\cos x + b\sin x as a single cosine or sine function.

    Recall the identity: acosx+bsinx=Rcos(xϕ),a\cos x + b\sin x = R\cos(x - \phi), where R=a2+b2R = \sqrt{a^2 + b^2} and ϕ\phi is an angle satisfying cosϕ=aR\cos \phi = \frac{a}{R} and sinϕ=bR\sin \phi = \frac{b}{R}.

    Here, a=3a = 3 and b=1b = 1, so: R=32+12=9+1=10.R = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}.

  2. Determine cosϕ\cos \phi and sinϕ\sin \phi: cosϕ=310,sinϕ=110.\cos \phi = \frac{3}{\sqrt{10}}, \quad \sin \phi = \frac{1}{\sqrt{10}}.

    Therefore: 3cosx+sinx=10(cosϕcosx+sinϕsinx).3\cos x + \sin x = \sqrt{10}\left( \cos \phi \cos x + \sin \phi \sin x \right).

    Using the cosine addition formula cos(AB)=cosAcosB+sinAsinB\cos(A - B) = \cos A \cos B + \sin A \sin B, we rewrite: 3cosx+sinx=10cos(xϕ).3\cos x + \sin x = \sqrt{10} \cos(x - \phi).

  3. Substitute into the original equation: The equation becomes: 10cos(xϕ)=1.\sqrt{10} \cos(x - \phi) = 1.

    Solving for cos(xϕ)\cos(x - \phi), divide both sides by 10\sqrt{10}: cos(xϕ)=110.\cos(x - \phi) = \frac{1}{\sqrt{10}}.

  4. Solve for xϕx - \phi: The general solution for cosθ=c\cos \theta = c is: θ=±arccos(c)+2nπ,nZ.\theta = \pm \arccos(c) + 2n\pi, \quad n \in \mathbb{Z}.

    Therefore: xϕ=±arccos(110)+2nπ.x - \phi = \pm \arccos\left(\frac{1}{\sqrt{10}}\right) + 2n\pi.

    Substituting back for ϕ\phi, where tanϕ=13\tan \phi = \frac{1}{3}, we can determine ϕ\phi: ϕ=arctan(13).\phi = \arctan\left(\frac{1}{3}\right).

  5. Write the final solution for xx: x=ϕ±arccos(110)+2nπ,nZ.x = \phi \pm \arccos\left(\frac{1}{\sqrt{10}}\right) + 2n\pi, \quad n \in \mathbb{Z}. Substituting ϕ=arctan(13)\phi = \arctan\left(\frac{1}{3}\right), the solution simplifies to: x=arctan(13)±arccos(110)+2nπ.x = \arctan\left(\frac{1}{3}\right) \pm \arccos\left(\frac{1}{\sqrt{10}}\right) + 2n\pi.


Final Notes:

The solution involves two angles:

  1. arctan(13)\arctan\left(\frac{1}{3}\right), which is ϕ\phi,
  2. arccos(110)\arccos\left(\frac{1}{\sqrt{10}}\right).

You can compute approximate numerical values for these angles using a calculator or software.


Let me know if you need further clarification or a step-by-step numerical approximation! 😊


Related Questions:

  1. How do you solve trigonometric equations using identities?
  2. What is the general solution of cosx=a\cos x = a?
  3. How do you express acosx+bsinxa\cos x + b\sin x as a single cosine function?
  4. How do you determine ϕ\phi if tanϕ=ba\tan \phi = \frac{b}{a}?
  5. What are the applications of trigonometric equations in real life?

Tip: Expressing sums like acosx+bsinxa\cos x + b\sin x as a single trigonometric function simplifies solving equations efficiently.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Cosine and Sine Functions

Formulas

aCos(x) + bSin(x) = R Cos(x - φ), where R = √(a² + b²)
Cos(A - B) = Cos(A)Cos(B) + Sin(A)Sin(B)
General solution for Cos(θ) = c: θ = ± arccos(c) + 2nπ

Theorems

Trigonometric addition formulas
Pythagorean theorem

Suitable Grade Level

Grades 11-12