〔中４〕兩條a.math題[寫明詳細步驟]

1. 3 sec2x + 6 tan2x = 4
3 (1 + tan2x) + 6 tan2x = 4
3 + 3tan2x + 6tan2x = 4
tan2x = 1/9
tanx = -1/3 (90 < x < 180, so tanx is negative)

Let y=1 and x=-3,
r = √[12 + (-3)2 ] = 10

So sinx = y/r = 1/√10

sinx + tanx = 1 / √10 - 1/3 = (3 - √10)/(3√10)

2. sin2x / 2 + 7cos2x = 1/9
sin2x + 14cos2x = 2/9
(1 - cos2x) + 14cos2x = 2/9
13cos2x = -7/9 ??? No solution

Should it be: sin2x / (2 + 7cos2x) = 1/9 ?
Then 9sin2x = 2 + 7cos2x
9 - 9cos2x = 2 + 7cos2x
16cos2x = 7
cos2x = 7 / 16
cosx = √7 /4 (270<x<360, so cosx is positive)

Let x=√7 and r=4,
y = -√ [ (4)2 - (√7)2 ] = -√9 = -3 (270<x<360, so y is negative)

tanx = y/x = -3/√7

1. 3(secX)^2+6(tanX)^2=4
3[(tanX)^2+1)+6(tanX)^2 = 4
9(tanX)^2 = 1
tanX = 1/3 (rejected as 90&lt;180, tanX &lt;0) or -1/3
taxX = -1/3 and sinX = 1/root(10) (90&lt;180, sinX&gt;0)
Hence sinX+tanX=1/root(10)-1/3 = (3-root(10))/3root(10)

2. (sinX)^2/2+7(cosX)^2 =1/9
(sinX)^2+14(cosX)^2 = 2/9
(sinX)^2+14[1-(sinX)^2] = 2/9
14-13(sinX)^2= 2/9
13(sinX)^2 = 124/9
(sinX)^2 = 124/117 (rejected, as 0&lt;=sinX^2&lt;=1)
NO SOLUTION
Is there any typo for Q2?