Trig question?

Question

I need help with this question...
The sum of the solutions of 2 sin^2 X= 7cos X - 2, 0 please show workings

Thanks Michael for your help but in my answers it say the answer is 2pi?

Answer 1

trigo identity: sin^2 x + cos^2 x = 1

therefore,

2sin^2 x = 2(1 - cos^2 x)

therefore,

2 - 2 cos^2 x = 7 cos x - 2

2cos^2 x + 7 cos x - 4 = 0

then, u will need to solve this like a normal quadratic equation, e.g. x^2 - 2x + 1 = 0

if u cant see it directly, try letting cos x be y

so, 2 y^2 + 7y - 4 = 0

then, u should get (2y-1)(y+4)=0

which means that y=1/2 or y=-4

therefore, cos x = 1/2 or -4
the answer cos x = -4 is rejected because it exceeds the limit of cos x, which should be between -1 and 1 inclusive.

so, since u know cos x = 1/2

meaning to say x = cos^-1 1/2
= pi/3 (in radian)

since ur range is 0 = 5pi/6
*lol, made a mistake.. kept thinking 2pi is only half a circle.

Answer 2

michael almost has it right, converting to a quadratic is the correct method, but the complete answer is
x = π/3 or x = 5π/3
These are the values for which cosx = 1/2 and 0 ≤ x ≤ 2π
You can always double check by substituting back into the original equation.

2 sin^2(π/3) = 7cos(π/3) -2
2(√3/2)^2 = 7(1/2) - 2
2(3/4) = 7/2 - 2
3/2 = 3/2
So π/3 works...

2 sin^2(5π/3) = 7cos(5π/3) -2
2(-√3/2)^2 = 7(1/2) - 2
2(3/4) = 7/2 - 2
3/2 = 3/2
So 5π/3 works...

And you can show that 2π doesn't work as well:

2sin^2(2π) = 7cos(2π) - 2
2(0)^2 = 7(1) - 2
2*0 = 5
0 = 5?

No, therefore, 2π does NOT work.

Answer 3

2(sin(x))^2 -7cos(x) = -2

2(1-(cos(x))^2) - 7cos(x) = -2

2 - 2 [cos(x)]^2 -7cos(x) = -2

-2[cos(x)]^2 - 7cos(x) + 4= 0, now you have a quadratic, therefore let z = cos(x)

-2z^2 -7z + 4 = 0

So just solve the quadratic and take the inverse cos of the solution(s) and look at the quadrant and you should have no trouble.

[ 7+/-sqrt(49 +4*2*4)]/-4, it is easy to mess this up, please check this. z could be -4 or z could be -2/-4 = 1/2, we can rule out -4 since cos cannot obtain that value

so what angle x makes cos(x) = 1/2?

Please check for errors.

Answer 4

2sin(x)^2 = 7cos(x) - 2
2(1 - cos(x)^2) = 7cos(x) - 2
2 - 2cos(x)^2 = 7cos(x) - 2
2cos(x)^2 + 7cos(x) - 4 = 0
(cos(x) + 4)(2cos(x) - 1) = 0

so

cos(x) + 4 = 0
cos(x) = -4
x = Undefined

so you have to go with

2cos(x) - 1 = 0
2cos(x) = 1
cos(x) = (1/2)
x = 60

x = (pi/3) or ((5pi)/3)

if you don't believe me, go to www.quickmath.com, click on Solve under Equations, type in 2sin(x)^2 = 7cos(x) - 2, click Solve and it will give you (pi/3) and (-pi/3), which is the same as saying (pi/3) and ((5pi)/3).

Answer 5

(Additional Details

3 minutes ago
Thanks Michael for your help but in my answers it say the answer is 2pi?)

Michel is right and your ans is wrong. if i put 2pi
LHS = 0 and rhs = 5 and they are not same.

Answer 6

2-2cos^2x=7cosx - 2
2cos^2x + 7cosx -4 = 0
(2cosx-1)(cosx +4) = 0
cosx=0.5, -4
cosx=-4 is not possible
x=arccos(0.5) = pi/3, 5pi/3

Answer 7

2sin^2x-7cosx+2=0
2(1-cos^2x)-7cosx+2=0
2-2cos^2x-7cosx+2=0
-2cos^2x-7cosx+4=0
=>2cos^2x+7cosx-4=0
=2cos^2x+8x-x-4=0
2cosx(cosx+4)-1(cosx+4)=0
(cosx+4)(2cosx-1)=0
2cosx=1
cosx=1/2
x=cos^-1(1/2)=+/-60*=>60* or 300*
=>pi/3 or5pi/3