Need help solving a trigonometric equation.

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December 28th 2010, 01:17 PM
paulaa

Need help solving a trigonometric equation.

thanks for the help :)
how will it be then if you have one like this 4 sin2(x) + 8 cos(x) = 7 ?
December 28th 2010, 01:20 PM
e^(i*pi)

If, as I assume, $\sin2(x) = \sin^2(x)$ then use the identity $\sin^2(x)+\cos^2(x) = 1$

Your equation will be $4\cos^2(x)-8\cos(x)+3=0$ which does factorise but it's probably easier to use the formula to solve for cos(x). You can then find x.
December 28th 2010, 01:23 PM
pickslides

Here's a kicker,

$4 \sin 2x + 8 \cos x = 7$

$4 \times 2 \sin x \cos x+ 8 \cos x = 7$

$8 \sin x \cos x+ 8 \cos x = 7$

$8 \cos x(\sin x+1) = 7$
January 14th 2011, 11:46 AM
harish21

Quote:

Originally Posted by paulaa

4sin²x + 8cosx = 7

sin²(x) + cos²(x) = 1 <=> sin²(x)=1-cos²(x)

replace sin²(x) with 1-cos²(x)

4(1-cos²(x)) + 8cos(x) - 7 = 0 <=> 4cos²(x) + 8cos(x) - 3 = 0You missed the negative sign before 4cos^2x

$4(1-cos²(x)) + 8cos(x) - 7 = 0$

$-4 \cos^2(x)+8\cos(x)-3=0$

$4\cos^2(x)-8\cos(x)+3=0$

move on..
January 14th 2011, 12:01 PM
paulaa

i saw that now, but it will be the same answer ?
January 14th 2011, 12:06 PM
paulaa

sorry i was wrong.

thank you : )
January 14th 2011, 01:04 PM
paulaa

i didn´t understod everything so i did it this way

4sin²x + 8cosx = 7

sin²(x) + cos²(x) = 1 <=> sin²(x)=1-cos²(x)

replace sin²(x) with 1-cos²(x)

4(1-cos²(x)) + 8cos(x) - 7 = 0 <=> 4cos²(x) + 8cos(x) - 3 = 0

replace cos(x) with t

4t² + 8t - 3 = 0

t² + 2t - 3/4 = 0

(t+1)² = 4/4 + 3/4 = 7/4

t1 = (7/4)^0.5 - 1

t2 = -(7/4)^0.5 - 1

x1 = cos^-1(((7/4)^0.5)-1)

but is this right to ?

Moderator edit: After deleting posts and moving things around, the ordering of how the posts should appear got messed up. For some of the responses in this thread to make sense, note that this post should come before post #4 (harish21's response).
January 14th 2011, 02:35 PM
skeeter

Quote:

Originally Posted by paulaa

i didn´t understod everything so i did it this way

4sin²x + 8cosx = 7

sin²(x) + cos²(x) = 1 <=> sin²(x)=1-cos²(x)

replace sin²(x) with 1-cos²(x)

4(1-cos²(x)) + 8cos(x) - 7 = 0 <=> 4cos²(x) + 8cos(x) - 3 = 0

4(1-cos²(x)) + 8cos(x) - 7 = 0 <=> 4 - 4cos²(x) + 8cos(x) - 7 = -4cos²(x) + 8cos(x) - 3 = 0

4cos²(x) - 8cos(x) + 3 = 0

[2cos(x) - 1][2cos(x) - 3] = 0

can you finish?
January 15th 2011, 10:44 AM
paulaa

the first equation:

i cant find the worong :/

sin(5x-((4x)/3) = 1/2

t=5x-((4x)/3)

sin(t)=1/2=sin(pi/6 +n*2pi)

5x - 4pi/3 = pi/6 + n*2pi

5x = 4pi/3 + pi/6 + n*2pi

5x = 9pi/6 + n*2pi

x = (9pi/6 + n*2pi)/5

x = 3pi/10 + n * 2pi/5

i think you will have 2 solutions will the other one be x = pi - 3pi/10 + n * 2pi/5

?