how to solve this trig equation?

Question

how does one solve this?

$\sin\left(x\right)+m\cos\left(x\right)=2m$

I've tried using a couple of formulas but nothing seems to work...

Is it even possible to solve explicitly for x ?

Answer 1

Hint:

$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Substitute in these formulae and multiply both sides with $e^{ix}$, you will obtain a quadratic equation in terms of $e^{ix}$.

Then you can easily solve for $x$.

(If no one else has given you a satisfying answer and I have time later I may provide a full solution, as there are some hidden traps which are indeed quite interesting.)

Answer 2

Let $\tan \alpha = 1/m$. Thus, $\cos \alpha = \dfrac{m}{\sqrt{1 + m^2}}$ and $\sin \alpha = \dfrac{1}{\sqrt{1 + m^2}}$. $$ \sin x + m\cos x = m \quad \Rightarrow \quad \dfrac{1}{\sqrt{1 + m^2}}\sin x + \dfrac{m}{\sqrt{1 + m^2}}\cos x = 2\dfrac{m}{\sqrt{1 + m^2}} \quad \Rightarrow $$ $$ \cos \alpha \cos x + \sin \alpha \sin x = 2\cos \alpha \quad \Rightarrow \quad \cos(x - \alpha) = 2\cos \alpha \quad \Rightarrow \quad x = \alpha + \cos^{-1}(2\cos \alpha) $$

Answer 3

1 :$\sin x +m \cos x=2m ⇒ \sin x =m (2-\cos x )$

2: $-1 ≤ \sin x ≤ 1 ⇒ -1 ≤ m(2 - \cos x ) ≤ 1 ⇒ 1 ≥ (\cos x -2) ≥ -1$

3: $-1 ≤ \cos x ≤ 1$

adding $-2$ to sides of inequality we get:

$-3 ≤ \cos x -2 ≤ -1$

To satisfy the second condition we must have $m=1/3$ so that:

$(1/3)(-3)=-1≤(1/3)(\cos x -2)≤(1/3)(-1) =-1/3<1$

Hence you have to solve following equation:

$\sin x +\frac{\cos x}{3}=\frac{2}{3}$

$x ≈ 20^o$

Answer 4

Use the tan-half-angle substitution ($t \rightarrow \tan\left( \frac{x}{2} \right)$) to turn the equation into a polynomial.

$$ \begin{cases} \sin(x) = \frac{2 t}{1+t^2} \\ \cos(x) = \frac{1-t^2}{1+t^2} \\ x =2 \tan^{-1}(t) \end{cases} $$

So now your equation becomes $$ \frac{2 t}{1+t^2} +m \frac{1-t^2}{1+t^2} = 2m$$ to be solved for $t$.

Finally, use $x = 2 \tan^{-1} (t)$ to get $x$ back.

$$ x = 2 \tan^{-1} \left( \frac{1 \pm \sqrt{1-3 m^2}}{3 m} \right) $$

Answer 5

Solution to the general $a \sin(x) +b \cos(x) = c$ equation.

1. Split $x$ into two variables $x = u+w$ and expand the sines & cosines $$ \cos(w) \left( b \cos(u) + a \sin(u) \right) + \sin(w) \left( a \cos(u) -b \sin(u) \right) = c $$
2. Set the first part equal zero by solving $ \cos(w)(b \cos(u) + a \sin(u)) = 0$ for $u$ $$ u = - \tan^{-1} \left(\frac{b}{a} \right) $$
3. Solve the remaining expression $\sin(w) \sqrt{a^2+b^2}=c$ for $w$ $$ w = \sin^{-1} \left( \frac{c}{\sqrt{a^2+b^2}} \right) $$
4. Assemble $x$

$$ \boxed{ x = \sin^{-1} \left( \frac{c}{\sqrt{a^2+b^2}} \right)- \tan^{-1} \left(\frac{b}{a} \right) }$$

or in this case

$$ x = \sin^{-1} \left( \frac{2 m}{\sqrt{1+m^2}} \right) -\tan^{-1}(m) $$

Answer 6

To solve the general equation $ a \sin(x) + b \cos(x) = c$ transform $a$ and $b$ into polar coordinates

$$ \begin{cases} a = r \cos(t) \\ b = r \sin(t) \end{cases} $$

Now you have

$$ r \sin(t) \cos(x) + r \cos(t) \sin(x) = r \sin(x+t) = c $$

$$ x = \sin^{-1} \left( \frac{c}{r} \right) -t $$

or

$$ \boxed{ x = \sin^{-1} \left( \frac{c}{\sqrt{a^2+b^2}} \right) - \tan^{-1} \left( \frac{b}{a} \right) } $$