Trigonometric equation $8\cos^3(x)-14\cos^2(x) \sin(x) + 5\cos(x) \sin^2(x) = 0$

Question

How would one go about solving

$$8\cos^3(x)-14\cos^2(x) \sin(x) + 5\cos(x) \sin^2(x) = 0$$

Should I try to factor it? Am I missing any essential formulas? Thanks.

Answer 1

Hint:

$$\cos(x) (8 \cos^2(x)-14\cos(x)\sin(x)+5\sin^2(x))=0$$

$$\cos(x) (2\cos x-\sin x)(4\cos x-5 \sin x)=0$$

Answer 2

Factoring does help,

$$8a^3-14a^2b+5ab^2$$

$$=a(8a^2-14ab+5b^2)$$

Now I think what multiplies to make $(8a^2)(5b^2)=40a^2b^2$ and adds to make $-14ab$. I know $(-10)(-4)=40$ and $-10+-4=-14$ so I proceed:

$$=a(8a^2-10ab-4ab+5b^2)$$

And factor by grouping etc..

Also if for example, $\sin x=2 \cos x$ then dividing both sides by $\cos x$ gives $\tan x=2$.

Answer 3

$$8\cos ^{ 3 } (x)-14\cos ^{ 2 } (x)\sin (x)+5\cos (x)\sin ^{ 2 } (x)=0\\ \cos { \left( x \right) \cdot \left( 8\cos ^{ 2 } (x)-14\cos { \left( x \right) } \sin (x)+5\sin ^{ 2 } (x) \right) =0 } \\ \cos { \left( x \right) =0,\quad } 8\cos ^{ 2 } (x)-14\cos { \left( x \right) } \sin (x)+5\sin ^{ 2 } (x)=0\\ { x }_{ 1 }=\frac { \pi }{ 2 } +2k\pi ,k\in \ Z ,\quad \quad 5\tan ^{ 2 }{ \left( x \right) } -14\tan { \left( x \right) } +8=0\quad \\ \left( 5\tan { \left( x \right) -4 } \right) \left( \tan { \left( x \right) -2 } \right) =0\\ \tan { \left( x \right) =\frac { 5 }{ 4 } } ,\tan { \left( x \right) =2 } \\ { x }_{ 2 }=\arctan { \frac { 5 }{ 4 } +k\pi ,\quad { x }_{ 3 }=\arctan { 2+k\pi , } } k\in Z\\ \quad \\ \\ \\ $$

Answer 4

• If $\cos x=0$, the equation is satisfied, whence a first series of solutions: $$x\equiv \frac\pi2\mod \pi.$$
• If $\cos x\ne 0$, note the equation is a homogeneous polynomial in $\cos x$ and $\sin x$. Dividing by $\cos^3x$, we obtain a quadratic polynomial in $\tan x$: $$8-14\tan x+5\tan^2x=0.$$ Its reduced discriminant is $\Delta'=9$, whence the solution $\tan x=2$, $\tan x=4/5$, and finally a second series of solutions in $x$: $$x\equiv \arctan 2,\quad x\equiv\arctan 4/5\mod\pi.$$