lượng giác thi thử chuyên phan bội châu lần 1

J

jet_nguyen

Bài 1:
Gợi ý:
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$$4\cos x -2\sin x- \cos2x =3$$$$\Longleftrightarrow 4\cos x -2\sin x- (\cos^2x-\sin^2) =3$$$$\Longleftrightarrow (\sin x-\cos x+1)(\sin x+\cos x+3)=0$$ Tới đây thì đơn giản rồi nhé.
 
J

jet_nguyen

Bài 2: Hình như đề thế này phải không bạn.
$$I=\int_0^ {\frac{\pi}{4}} \dfrac{2x+\cos^2x}{1+\sin 2x} \mathrm{d} x.$$
Ta có:
$$I=\int_0^ {\frac{\pi}{4}} \dfrac{2x+\cos^2x}{1+\sin 2x} \mathrm{d} x.$$$$=\int_0^ {\frac{\pi}{4}} \dfrac{4x+1+\cos2x}{2(1+\sin 2x)} \mathrm{d} x.$$$$=\int_0^ {\frac{\pi}{4}} \dfrac{\cos2x}{2(1+\sin 2x)} \mathrm{d} x+\int_0^ {\frac{\pi}{4}} \dfrac{4x+1}{2(1+\sin 2x)} \mathrm{d} x.$$$$=\int_0^ {\frac{\pi}{4}} \dfrac{1}{4(1+\sin 2x)} d\sin2x+\int_0^ {\frac{\pi}{4}} \dfrac{4x+1}{4\cos^2\left(x-\dfrac{\pi}{4}\right)} \mathrm{d} x.$$$$=\dfrac{\ln|\sin2x+1|}{4}+\dfrac{4x+1}{4}.tan\left(x-\dfrac{\pi}{4}\right)\bigg|_0^ {\frac{\pi}{4}} -\int_0^ {\frac{\pi}{4}} \tan\left(x-\dfrac{\pi}{4}\right)dx.$$$$=\dfrac{\ln|\sin2x+1|}{4}+\dfrac{4x+1}{4}.\tan\left(x-\dfrac{\pi}{4}\right) +\ln|\cos\left(x-\dfrac{\pi}{4}\right)|\bigg|_0^ {\frac{\pi}{4}}$$
 
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