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CIE 9709 2023 June Paper 13 Q4

A Level / CIE / P1

CIE 9709 2023 June Paper 13 Paper · Question 4

题目

Problem

(a) Show that the equation

3tan2x3sin2x4=03\tan^2 x - 3\sin^2 x - 4 = 0

may be expressed in the form acos4x+bcos2x+c=0a\cos^4 x + b\cos^2 x + c = 0, where aa, bb and cc are constants to be found.

[3]

(b) Hence solve the equation 3tan2x3sin2x4=03\tan^2 x - 3\sin^2 x - 4 = 0 for 0x1800^\circ \leq x \leq 180^\circ.

[4]
题目中文翻译

(a) 证明方程

3tan2x3sin2x4=03\tan^2 x - 3\sin^2 x - 4 = 0

可以表示成 acos4x+bcos2x+c=0a\cos^4 x + b\cos^2 x + c = 0 的形式,其中 aabbcc 是需要求出的常数。

[3]

(b) 由此解方程 3tan2x3sin2x4=03\tan^2 x - 3\sin^2 x - 4 = 0,其中 0x1800^\circ \leq x \leq 180^\circ

[4]

解答