1) a) cos²x=1/2
⇔ (1+cos2x)/2 = 1/2
⇔ cos2x = 0
⇔ x = π/4 +k.π/2 (k∈Z)
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b) sin²(x-π/4)=cos²x
⇔ $\frac{1-cos(x-π/2)}{2}$ = $\frac{1+cos2x}{2}$
⇔ $\frac{1-sin2x}{2}$ = $\frac{1+cos2x}{2}$
⇔ sin2x+cos2x = 0
⇔ (√2) . sin(2x+π/4) =0
⇔ 2x+π/4 =kπ
⇔ x = -π/8 +k.π/2 (k∈Z)
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c) cos²3x +sin²2x=1
⇔ cos²3x = 1 - sin²2x
⇔ cos²3x = cos²2x
⇔ $\frac{1+cos6x}{2}$ = $\frac{1+cos4x}{2}$
⇔ cos6x - cos4x = 0
⇔ -2sin5x.sinx=0
⇔ \(\left[ \begin{array}{l}sin5x=0\\sinx=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=kπ/5\\x=kπ\end{array} \right.\) (k∈Z)
⇔ x=kπ/5 (k∈Z)
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d) cos7x.cosx = cos5x.cos3x
⇔ 1/2 .(cos8x+cos6x) = 1/2 .(cos8x+cos2x)
⇔ cos6x - cos2x = 0
⇔ -2sin4x.sin2x = 0
⇔ \(\left[ \begin{array}{l}sin4x=0\\sin2x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=kπ/4\\x=kπ/2\end{array} \right.\) ⇔ x=kπ/4 (k∈Z)
