solve sec^4 (2x) = 4, sin 2(x) + sin4(x) = 0

[bstevens]

I have homework problems that I am completely stuck on. Please help!

1. solve sec^4 (2x) = 4 for solutions in the interval (0, 2pi)

2. Solve sin 2(x) + sin4(x) = 0 for solutions in the interval (0, 360 degrees)

 

[stapel]

1) Hint: Take the fourth root of each side.

2) Do you mean "sin(2x) + sin(4x) = 0", or something else?

Thank you.

Eliz.

 

[tkhunny]

1) Hint: Take the fourth root of each side.

You're slipping, Eliz. That't what I may have said.

[sec(2x)]^4 - 4 = 0

([sec(2x)]^2 + 2)([sec(2x)]^2 - 2) = 0

([sec(2x)]^2 + 2)(sec(2x) + sqrt(2))(sec(2x) - sqrt(2)) = 0

[sec(2x)]^2 + 2 = 0 -- Nothing there, but the other 2...

You can find all 8 if you look at all the factors.

 

[stapel]

You're slipping, Eliz....

Okay; take the second root twice:

. . . . .sec⁴(2x) = 4

. . . . .sec²(2x) = ­­­± 2

Since sec²(2x) cannot be negative, then:

. . . . .sec²(2x) = 2

. . . . .sec(2x) = ± sqrt[2]

And so forth.

Eliz.

 

[soroban]

Hello, bstevens!

I would guess that #2 has exponents.

2. Solve: $\displaystyle \,\sin^2x\,+\,\sin^4x\:=\:0$ on the interval $\displaystyle (0^o,\,360^o)$

Factor: $\displaystyle \,\sin^2x\left(1\,+\,\sin^2x\right)\:=\:0$


Set each factor equal to zero and solve.

. . \(\displaystyle \sin^2x\,=\,0\;\;\Rightarrow\;\;\sin x\,=\,0\;\;\Rightarrow\;\;\fbox{x\:=\:0^o,\,180^o,\,360^o}\)

. . $\displaystyle \sin^2x\,+\,1\:=\:0\;\;\Rightarrow\;\;\sin^2x\,=\,-1\;\;\Rightarrow\;\;\sin x \,=\,\pm\sqrt{-1}\;$ . . . no real roots