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Solve the trigonometric equation:
sin 4x = – 2 · sin 2x
sin 4x + 2 · sin 2x = 0 (i)
Use the double-angle formula for sin 4x:
sin 4x = 2 · sin 2x · cos 2x
Substitute that into (i), and you have
2 · sin 2x · cos 2x + 2 · sin 2x = 0 (ii)
Now, take out the common factor 2 · sin 2x:
2 · sin 2x · (cos 2x + 1) = 0
sin 2x · (cos 2x + 1) = 0
If the product above equals zero, then either sin 2x or (cos 2x + 1) must be zero. So,
sin 2x = 0 or cos 2x + 1 = 0
sin 2x = 0 or cos 2x = –1
Solving them separately:
• sin 2x = 0
2x = k · π
k · π
x = ———
2
where k is an integer (all integer multiples of π/2).
• cos 2x = – 1
2x = π + k · 2π
2x = (1 + 2k) · π
(1 + 2k) · π
x = ——————
2
where k is an integer (all odd integer multiples of π/2).
But notice that the solution for sin 2x = 0 also includes those particular cases that we found when solving cos 2x = – 1.
Therefore, the solution set is simply
[tex]\mathsf{S=\left\{x\in\mathbb{R}:~~x=\dfrac{k\cdot \pi}{2},~~k\in\mathbb{Z}\right\}}[/tex]
(all integer multiples of π/2).
I hope this helps. =)
Tags: solve trigonometric equation double angle formula factor sine cosine sin cos trig trigonometry
