solve sin2x+2cos2x=1?

Question

from 0

Answer 1

sin2x+2cos2x=1
=>2sinxcosx+4cos^2[x]-2=1
=>2cos^2[x]+sinxcosx=3/2
=>cosx(2cosx+sinx)=3/2
=>2+tanx=3sec^2[x]/2 dividing across by cos^2[x]
=>2+tanx=3(1+tan^2[x])/2
=>3tan^2[x]-2tanx-1=0
=>(3tanx+1)(tanx-1)=0
=>tanx=-1/3 or tanx=1
First set of answers lie in the 2nd and 4th quadrants of the unit circle since tangent is negative here and the second set of answers lie in the 1st and 3rd quadrants.
Using my calculator i find the first two to be 161.565degrees and 341.465degrees, with the second set being 45degrees and 225degrees. Therefore the answers are 45degs, 161.565degs, 225degs. & 341.565degs......HOPE THATS HELPED!!!

Answer 2

sin2x + sin4x = 0 sin2x + 2sin2xcos2x = 0 sin2x(a million + 2cos2x) = 0 sin2x = 0 or cos2x = -a million/2 sin2x = 0 ==> 2x = k? ==> x = k?/2 cos2x = -a million/2 ==> 2x = ±2?/3 + ok.2? ==> x = ±?/3 + k? The suggestions interior the area [0,2?) are 0, ?/2, ?, 3?/2, ?/3, 4?/3, 2?/3, 5?/3

Answer 3

sin2x + 2cos2x = 1
Let say y=2x
siny + 2cosy = 1
siny + 2(1-sin^2 y)^(1/2) = 1
(1-sin^2 y)^(1/2)=1/2(1-siny)
Square both sides
1-sin^2 y = 1/4(1-siny)^2
1-sin^2 y = 1/4(1+sin^2 y-2siny)
5sin^2 y -2sin y -3=0
(5sin y+3)(sin y-1)=0
sin y = -3/5 or sin y = 1

since 0 0 0
And
both siny and cosy cannot -ve
at same time.

So, 0 3pi/2 7pi/2
sin y = -3/5
y = 323.13deg, 683.13deg
x=161.565deg, 341.565deg

sin y = 1
y = 90deg, 450deg
x = 45deg, 225deg

Answer 4

2sinxcosx+2(1-2sin^2x)=1
2sinxcosx+2-4sin^2x=1
2sinxcosx-4sin^2x=-1
2sinxcosx-4sin^2x=-(sin^2x+cos^2x) because 1=cos^2x+sin^2x
move everything from left sides to the right side and combine like terms.
3sin^2x-2sinxcosx-cos^2x=0
because x=pi/2 is not a solution so we divided both sides by cos^2x.
3tan^2-2tanx-1=0
that is quaratic for tanx
we factor( tanx-1)(3tanx+1)=0. therefore tanx-1=0 and 3tanx+1=0
so tanx=1 and tanx=-1/3
the solutions are x=pi/4 (45 degree) , x=3pi/4(225 degree in the third quadrant) and x=tan^-1(-1/3) remember tan^-1 is inverse of tanx. sorry for any inconvenience because i cant write square of sinx of cosx and you will see a mistake cos... that is cosine square of x. sorry. i hope that may be helping you a little bit.good luck

Answer 5

sin 2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2
(sinx)^2 + (cosx)^2 = 1
therefore the given eq can be reduced to
3(sinx)^2 - 2sinxcosx - (cosx)^2= 0
solution is
3sinx = -cosx ie tanx = -1/3
or
sinx = cosx ; ie tan x = 1
now u can continue it urself
good luck.

Answer 6

hey dude, i can make up things too