How to solve this A-level pure math 1 question

[From: ] [author: ] [Date: 13-09-25] [Hit: ]

150°, 390°, 510°,.........

so, (2sinx - 1)(sinx + 2) = 0

=> sinx = 1/2 or sinx = -2....not possible as -1 ≤ sinx ≤ 1

Then, x = 30°, 150°, 390°, 510°,.....

Now, the equation in the second part is the same as in the first part but with x = 2y

so, 2y = 30°, 150°, 390°, 510°,....e.t.c.

i.e. y = 15°, 75°, 195°, 255°,....

=> y = 15° and 75° in the domain 0° ≤ y ≤ 180°

:)>

-

2cosx = 3tanx
2cosx = 3sinx/cosx
2cos²x = 3sinx
2(1-sin²x) = 3sinx
2sin²x + 3sinx - 2 =0
let sinx = t

2t² +3t - 2=0
(2t-1)(t+2) = 0
t = 1/2 or t = -2
since -1 t = 1/2 only
sinx = 1/2
sin2y = 1/2
sin2y = sin30
2y = 30
y = 15

by general solution for any integer n
2y = n(180) +(-1)^n 30
y = 90n + (-1)^n15
when n=1
y = 90 - 15 = 75
when n = 2
y = 180 - 15 = 165

so ans 15,75,165 degrees

-

2 cos 2y = 3 tan 2y
2 cos 2y = 3 sin 2y / cos 2y

2 cos^(2y) = 3 sin 2y

2( 1 - sin^2(2y) ) = 3 sin 2y

you have a quadratic in sin(2y) put sin(2y) = x

2(1 -x^2) = 3x

2x^2 + 3x -2 = 0

x = -3 +/- sqrt( 9 + 4*2*2) /4

x = 1/2 or -2 -2 for sin not a valid value

sin(2y) = 1/2

solve it

-

2 cos 2y = 3 tan 2y
2 cos^2 2y = 3 sin 2y
2 (1 -- sin^2 2y) = 3 sin 2y
2 sin^2 2y + 3 sin 2y -- 2 = 0
(2 sin 2y -- 1)(sin 2y + 2) = 0 giving sin 2y = --2 (rejected) and
sin 2y = 1/2 = sin pi/6 whence y = npi/2 + (--1)^n (pi/12)
y = pi/12, 5pi/12 ANSWER

-

2cosx=3tanx
2cosx=3sinx/cosx
2cos^2x -3sinx=0
2(1 - sin^x)-3sinx=0
2 - 2sin^2x_3sinx=0
2sin^2x+3six-2=0

12

keywords: level,this,solve,math,How,question,pure,to,How to solve this A-level pure math 1 question