Solving Trigonometric Eqn: 4sin^2(x) cos(x) = tan^2(x)

[ninatemple]

• 4sin^2(x) cos(x) = tan^2(x)

 

[ninatemple]

Solve Trig Eqtn on 0º ≤ θ ≤ 360º: 4sin^2(x) cos(x) = tan^2(x)

• 4sin^2(x) cos(x) = tan^2(x)

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My working went like this:
4[1-cos^2(x)] cos(x) = tan^2(x)
4-4cos^2(x) cos(x) - tan^2(x) = 0

then take out cos(x) bc it's a common factor?

 

[stapel]

4sin^2(x) cos(x) = tan^2(x)

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My working went like this:
4[1-cos^2(x)] cos(x) = tan^2(x)
4-4cos^2(x) cos(x) - tan^2(x) = 0

then take out cos(x) bc it's a common factor?

Where are you getting that cos(x) is a common factor of 4, 4cos³(x), and tan²(x)?

Instead, in the usual manner, let's try starting with your substitution, and then multiplying everything out, and converting everything to sines and cosines:

. . . . .\(\displaystyle 4\, \sin^2(x)\, \cos(x)\, =\, \tan^2(x)\)

. . . . .\(\displaystyle 4\, \left[\, 1\, -\, \cos^2(x)\, \right]\, \cos(x)\, =\, \dfrac{\sin(x)}{\cos(x)}\)

. . . . .\(\displaystyle \left[\, 4\, -\, 4\, \cos^2(x)\, \right]\, \cos(x)\, =\, \dfrac{\sin(x)}{\cos(x)}\)

. . . . .\(\displaystyle 4\, \cos(x)\, -\, 4\, \cos^2(x)\, \cos(x)\, =\, \dfrac{\sin(x)}{\cos(x)}\)

. . . . .\(\displaystyle 4\, \cos(x)\,-\, 4\, \cos^3(x)\, =\, \dfrac{\sin(x)}{\cos(x)}\)

Now, noting that we can't use any x-value for which the cosine equals zero (because this would cause division by zero, which is not allowed), let's multiply through by cos(x):

. . . . .\(\displaystyle 4\, \cos^2(x)\, -\, 4\, \cos^4(x)\, =\, \sin(x)\)

Can you see a way to convert this all into sines, and then try solving?

 

[Ishuda]

• 4sin^2(x) cos(x) = tan^2(x)

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My working went like this:
4[1-cos^2(x)] cos(x) = tan^2(x)
4-4cos^2(x) cos(x) - tan^2(x) = 0

then take out cos(x) bc it's a common factor?

Hint
\(\displaystyle 4\, sin^2(x)\, cos(x)\, =\, tan^2(x)\, =\, \frac{sin^2(x)}{cos^2(x)}\)