Giải phương trình dùng công thức lượng giác biến đổi về phương trình cơ bản:
a)cos²x-cos²(pi/2-x)=1/2
b)cos⁴x-sin⁴x=✓3/2
c)cos⁴x+sin⁴x=3/4
d)4sinx.sin(pi/3-x).sin(pi/3+x)=✓3/2
e)4cosx.cos(pi/3-x).cos(pi/3+x)=✓2/2
Giải phương trình dùng công thức lượng giác biến đổi về phương trình cơ bản:
a)cos²x-cos²(pi/2-x)=1/2
b)cos⁴x-sin⁴x=✓3/2
c)cos⁴x+sin⁴x=3/4
d)4sinx.sin(pi/3-x).sin(pi/3+x)=✓3/2
e)4cosx.cos(pi/3-x).cos(pi/3+x)=✓2/2
$\begin{array}{l}a)\,\cos^2x – \cos^2\left(\dfrac{\pi}{2} – x\right) = \dfrac{1}{2}\\ \Leftrightarrow \cos^2x – \sin^2x = \dfrac{1}{2}\\ \Leftrightarrow \cos2x = \cos\dfrac{\pi}{3}\\ \Leftrightarrow 2x = \pm \dfrac{\pi}{3} + k2\pi\\ \Leftrightarrow x = \pm \dfrac{\pi}{6} + k\pi \quad (k \in \Bbb Z)\\ b)\,\cos^4x – \sin^4x = \dfrac{\sqrt3}{2}\\ \Leftrightarrow (\cos^2x – \sin^2x)(\cos^2x + \sin^2x) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow \cos2x = \cos\dfrac{\pi}{6}\\ \Leftrightarrow 2x = \pm \dfrac{\pi}{6} + k2\pi\\ \Leftrightarrow x = \pm \dfrac{\pi}{12} + k\pi \quad (k \in \Bbb Z)\\ c)\,\cos^4x + \sin^4x = \dfrac{3}{4}\\ \Leftrightarrow (\cos^2x + \sin^2x)^2 – 2\sin^2x\cos^2x = \dfrac{3}{4}\\ \Leftrightarrow 1 – \dfrac{1}{2}(2\sin x\cos x)^2 = \dfrac{3}{4}\\ \Leftrightarrow \dfrac{1}{2}\sin^22x = \dfrac{1}{4}\\ \Leftrightarrow 1 – \cos4x = 1\\ \Leftrightarrow \cos4x = 0\\ \Leftrightarrow 4x = \dfrac{\pi}{2} + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{8} + k\dfrac{\pi}{4} \quad (k \in \Bbb Z)\\ d)\,4\sin x.\sin\left(\dfrac{\pi}{3} – x\right).\sin\left(\dfrac{\pi}{3} + x\right) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow 4\sin x.\left[-\dfrac{1}{2}\left(\cos\dfrac{2\pi}{3} – \cos2x\right)\right]=\dfrac{\sqrt3}{2}\\ \Leftrightarrow 4\sin x.\left[\dfrac{1}{2}\left(\dfrac{1}{2} + \cos2x\right)\right]=\dfrac{\sqrt3}{2}\\ \Leftrightarrow 4\sin x.\left(\dfrac{2\cos2x + 1}{4} \right)=\dfrac{\sqrt3}{2}\\ \Leftrightarrow 2\sin x\cos2x + \sin x =\dfrac{\sqrt3}{2}\\ \Leftrightarrow 2.\dfrac{1}{2}(\sin3x – \sin x) + \sin x = \dfrac{\sqrt3}{2}\\ \Leftrightarrow \sin3x = \sin\dfrac{\pi}{3}\\ \Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{3} + k2\pi\\3x = \dfrac{2\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\\x = \dfrac{2\pi}{9} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\\ e)\,4\cos x.\cos\left(\dfrac{\pi}{3} – x\right).\cos\left(\dfrac{\pi}{3} + x\right) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow 4\cos x.\left[\dfrac{1}{2}\left(\cos\dfrac{2\pi}{3} + \cos2x\right)\right]= \dfrac{\sqrt3}{2}\\ \Leftrightarrow 4\cos x.\left[\dfrac{1}{2}\left(\cos2x – \dfrac{1}{2}\right)\right]= \dfrac{\sqrt3}{2}\\ \Leftrightarrow 4\cos x.\left(\dfrac{2\cos2x – 1}{4}\right) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow 2\cos x\cos2x – \cos x = \dfrac{\sqrt3}{2}\\ \Leftrightarrow 2.\dfrac{1}{2}(\cos3x + \cos x) – \cos x =\dfrac{\sqrt3}{2}\\ \Leftrightarrow \cos3x = \cos\dfrac{\pi}{6}\\ \Leftrightarrow 3x = \pm \dfrac{\pi}{6} + k2\pi\\ \Leftrightarrow x = \pm \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\quad (k \in \Bbb Z) \end{array}$