Can $\sin^5 x + \cos^3 x$ Equal 1?

  • Context: High School 
  • Thread starter Thread starter anemone
  • Start date Start date
Click For Summary
SUMMARY

The equation $\sin^5 x + \cos^3 x = 1$ has been successfully solved by forum members Opalg, kaliprasad, and lfdahl. The solutions provided demonstrate that the equation holds true for specific values of \(x\) where the sine and cosine functions align to satisfy the equation. The discussion emphasizes the importance of understanding trigonometric identities and their applications in solving polynomial equations involving sine and cosine.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine and cosine.
  • Familiarity with polynomial equations and their solutions.
  • Knowledge of trigonometric identities and their properties.
  • Basic skills in mathematical proof techniques.
NEXT STEPS
  • Explore the derivation of trigonometric identities relevant to polynomial equations.
  • Study the graphical representation of sine and cosine functions to visualize solutions.
  • Learn about solving higher-degree polynomial equations involving trigonometric functions.
  • Investigate numerical methods for finding roots of trigonometric equations.
USEFUL FOR

Mathematicians, students studying trigonometry, and anyone interested in solving complex equations involving trigonometric functions.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----

Find all the real solutions of the equation $\sin^5 x+\cos^3 x=1$.

-----

 
Physics news on Phys.org
Congratulations to the following members for their correct solution!(Cool)

1. Opalg
2. kaliprasad
3. lfdahl

Solution from Opalg:
Let $f(x) = \sin^5x+ \cos^3x - 1$. Since $f$ has period $2\pi$ it will be enough to find solutions of $f(x)=0$ in the interval $0\leqslant x < 2\pi$.

If $\sin x$ or $\cos x$ is negative then $f(x)<0$. That rules out the interval $\pi/2 <x < 2\pi$. So it will be enough to find solutions of $f(x)=0$ in the interval $0\leqslant x \leqslant \pi/2$.

If $x=0$ or $x=\pi/2$ then $f(x)=0$. That leaves the interval $0<x<\pi/2$. But in that interval $0<\sin x <1$ and $0<\cos x <1$. Therefore $ \sin^5x+ \cos^3x < \sin^2x+ \cos^2x =1$ and so $f(x) <0$.

Thus the only solutions are $x = 2k\pi$ and $x = \left(2k+\frac12\right)\pi$ for $k\in\Bbb{Z}$.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K