Solution for the equation $cos⁡(40+theta) = 3sin⁡(50+theta)$

By multiplying the left-hand side with $sin 20$ both up and down, we get $dfrac{sin 20^circ}{sin 20^circ} cdot cos 20^circ cos 40^circ cos 80^circ = dfrac{sin 40^circ}{2sin 20^circ} cdot cos 40^circ cos 80^circ = dfrac{sin 80^circ}{4sin 20^circ} cdot cos 80^circ = dfrac{sin 160^circ}{8sin 20^circ}=dfrac{sin 20^circ}{8sin theta_1}$. Instead of squaring which introduces extraneous roots, we can write $9=rsin y$ and $40=rcos y$, where $r>0$, and then add and square to get $r^2=41^2implies r=41$.


Question:

Find the solution for the given equation within the range of $0≤theta≤180$:
$$
cos⁡(40+theta) = 3sin⁡(50+theta)
$$
Note that $cos⁡(40)$ is equivalent to $sin(⁡50)$.


Solution

The equation $cos(40+theta) = 3sin(50+theta)$ can be written as $cos(40)costheta -sin(40)sintheta = 3(sin(50)costheta +sintheta cos(50))$. Alternatively, it can be expressed as $sin(50)costheta-cos(50)sintheta = 3(sin(50)costheta+sintheta cos(50))$. Simplifying this further gives $2sin(50)costheta+4sintheta cos(50) = 0$.


Solution 1:


Solution 2:


Note that the equation $cos(theta)=0$ does not yield any solution. By rearranging the last line with the help of comments, we obtain $tan(-50)=2tan(theta)$. It is important to remember that $tan(-150)=dfrac1{sqrt 3}$ and the trigonometric identity $tan(3phi)=dfrac{3tan(phi)-tan^3(phi)}{1-3tan^2(phi)}$. Therefore, we can express $dfrac{6tan(theta)-8tan^3(theta)}{1-12tan^2(theta)}$ as $dfrac1{sqrt 3}$. This leads to the quadratic equation $dfrac1{sqrt 3}-6tan(theta)-4sqrt3tan^2(theta)+8tan^3(theta)=0$ in terms of $tan(theta)$.

The equation $dfrac1{8sqrt 3}-dfrac34tan(theta)-dfrac{sqrt3}2tan^2(theta)+tan^3(theta)=0$ can be transformed by letting $x+dfrac{sqrt3}6=tan(theta)$. This results in the equation $f(x)=x^3-x-dfrac{sqrt3}9=0$.

The discriminant of the function $f(x)$ yields $delta^2=3$ using the Cardano formula. This is found by substituting the given values to get $delta^2=-4(-1)^3-27(-dfrac{sqrt3}9)^2$. Thus, one of the roots of $f$ is $chi=sqrt[3]{dfrac{sqrt3}{18}+dfrac{sqrt{-1}}6}+sqrt[3]{dfrac{sqrt3}{18}-dfrac{sqrt{-1}}6}$.

Note that $f(frac{sqrt{3}}{3}) > 0$, $f(1) < 0$, and $f(2) > 0$. By the intermediate value theorem, $f$ has a minimum of two real roots. It is important to mention that since $f(x)in Bbb R[x]$, both $alpha$ and its complex conjugate $bar{alpha}nealpha$ are non-real roots of $f$. As a result, $f$ has three real roots. This leads us to the conclusion that $chi=sqrt[3]{frac{sqrt3}{18}+frac{sqrt{-1}}{6}}+sqrt[3]{frac{sqrt3}{18}-frac{sqrt{-1}}{6}}in Bbb R$.

Start by defining $p$ and $q$ as:
$p=sqrt[3]{dfrac{sqrt3}{18}+dfrac{sqrt{-1}}6}$ and $q=sqrt[3]{dfrac{sqrt3}{18}-dfrac{sqrt{-1}}6}$.
It follows that $chi=p+q$, and thus, $chi^3=(p+q)^3=p^3+q^3+3pq(p+q)$.
Additionally, we have $p^3+q^3=dfrac {sqrt 3}9$ and $pq=sqrt[3]{(dfrac{sqrt 3}{18})^2+dfrac 1{36}}$.
Assuming that $p+q<0$, we get $|p+q|^3lt 3pq|p+q|$, which implies $(p+q)^2=p^2+2pq+q^2lt 3pq implies p^2+q^2lt pq$.
However, applying the A.M.-G.M. inequality, we get $p^2+q^2gt 2|pq|=2pq$, which gives us $0lt 2pqlt p^2+q^2lt pq$, a contradiction.

Given that $chi=p+qgt 0$, we have $tan(theta)=chi+dfrac{sqrt3}6$. This can also be expressed as $tan(theta)=sqrt[3]{dfrac{sqrt3}{18}+dfrac{sqrt{-1}}6}+sqrt[3]{dfrac{sqrt3}{18}-dfrac{sqrt{-1}}6}+dfrac{sqrt3}6gt 0$. As $arctan(phi)gt 0$ if and only if $phi gt 0$, we can conclude that $theta=arctan(sqrt[3]{dfrac{sqrt3}{18}+dfrac{sqrt{-1}}6}+sqrt[3]{dfrac{sqrt3}{18}-dfrac{sqrt{-1}}6}+dfrac{sqrt3}6) gt 0$. Therefore, $theta$ is the desired solution.

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