116k views
4 votes
Solve 8cos(2w)=8sin²(w)+4 for all solutions 0≤w<2π

User Daxelrod
by
8.3k points

1 Answer

0 votes

Final answer:

The equation 8cos(2w)=8sin²(w)+4 is solved using trigonometric identities and algebra. After simplifying and solving for sin(w), we find the solutions for w within the range 0≤w<2π, yielding approximately 0.41, 2.73, 3.55, and 5.87 radians.

Step-by-step explanation:

To solve the equation 8cos(2w)=8sin²(w)+4, we can use trigonometric identities to simplify the equation. Since cos(2w) is an identity that can be written as 1-2sin²(w), we can rewrite the original equation as:

8(1-2sin²(w))=8sin²(w)+4.

Expanding and simplifying this equation will give us:

8-16sin²(w)=8sin²(w)+4

24sin²(w)=4

sin²(w)=\(\frac{1}{6}\)

Taking the square root on both sides gives us:

sin(w)=\(\pm\sqrt{\frac{1}{6}}\)

Now we find all solutions for w within the range of 0≤w<2π. We know the sin function is positive in the first and second quadrants and negative in the third and fourth quadrants.

The solutions are approximately:

  • w≈ 0.41, 2.73 radians (for positive square root)
  • w≈ 3.55, 5.87 radians (for negative square root)

User Evan Pu
by
8.3k points

Related questions