Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

Question

Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

My Try

$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!

Answer 1

$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

$6\sin{x}\cos{x} -9\sin^2{x} - \cos^2{x} -2\cos{x} + 6\sin{x} - 1=0$

$-(3 \sin x - \cos x)^2 + 2(3 \sin x - \cos x) - 1= 0$

$(3 \sin x - \cos x)^2 - 2(3 \sin x - \cos x) + 1= 0$

Answer 2

$x=2y$ we get after some manipulation $$-4\cos^4 y-36\cos^2y\sin^2 y+24\cos^3 y\sin y=0$$ $$-4\cos^2 y{(\cos y-3\sin y)}^2=0$$ can you end it??

Answer 3

Abbreviating $\sin x$ and $\cos x$ to $s$ and $c$, and writing $\sin2x=2sc$ and $\cos2x=2c^2-1$, the equation becomes

$$6sc+4(2c^2-1)-2c+6s-6=0$$

Factoring out a $2$ and grouping terms, we rewrite this as

$$3s(c+1)+4c^2-c-5=0$$

But $4c^2-c-5=(4c-5)(c+1)$, so we have two solutions in $s$ and $c$:

$$c=-1\quad\text{and}\quad3s+4c=5$$

Reverting to $\sin x$ and $\cos x$ and recognizing the Pythagorean triple $3^2+4^2=5^2$, these become

$$\cos x=-1\quad\text{and}\quad\sin(x+\theta)=1$$

where $\sin\theta=4/5$ and $\cos\theta=3/5$, e.g. $\theta=\arcsin(4/5)$. So the solution set is

$$\{\pi+2n\pi\mid n\in\mathbb{Z}\}\cup\{\pi/2-\arcsin(4/5)+2n\pi\mid n\in\mathbb{Z} \}$$

Remark: The abbreviations $s$ and $c$ are merely a convenience. Once I get all the sines and cosines down to $\sin x$ and $\cos x$, I find it tedious to write the functions out in full over and over again.

Answer 4

Let $t = \tan (x/2)$. Then $\sin x = 2t/(1+t^2)$ and $\cos x = (1-t^2)/(1+t^2)$. Your equation looks quartic after multiplying both sides by $(1+t^2)^2$. $$12t(1-t^2)+4((1-t^2)^2-4t^2)-(2-2t^2-12t)(1+t^2)-6(1+t^2)^2$$ but it simplifies to $-4(3t-1)^2=0$, so $t = 1/3$, giving $x = 2(\tan^{-1} t + n\pi) = 2(\tan^{-1}{\frac13}+n\pi)$ for any integer $n$, since $\tan$ has period $\pi$.

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The above $t$-substitution omits the possibility that $x = (2n+1) \pi$, which is actually a solution: cross out the $\sin$ in the equation and observe that $4(1)-2(-1)-6 = 0$.

Answer 5

Like Barry Cipra,

replacing $\sin x=s,\cos x=c,\cos2x=2c^2-1,\sin2x=2sc$

$$3(2sc)+4(3c^2-1)-2c+6s-6=0$$

$$\iff4c^2-c(1-3s)+3s-5=0$$

$$\implies c=\cdots=-1,\dfrac{5-3s}4$$

If $\cos x=c=-1, x=(2n+1)\pi$

Else $4\cos x+3\sin x=5$

Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$

See also: Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$