Question

35sinx-100cosx+150cos2x=0 , i need to get values of x.

Answer 1

sorry is it
\[\cos(2x)\]
or
\[(\cos x)^{2}\]?

Answer 2

cos(2x)

Answer 3

I rewrite your equation as below:
\[7 \sin x-20 \cos x +30 (cosx)^{2}-30 (\sin x)^{2}=0\]
from which:
\[\cos x[(7 \tan x-20)+30 \cos x(1-(\tan x)^{2}]=0\]

Answer 4

Oh you divided everything by 5 first.

Answer 5

now, applying the canceling law of product, we have:
cos x=0
or\[(7 \tan x-20)+30 \cos x (1-(\tan x)^{2}))=0\]

Answer 6

@Jhannybean that's right!

Answer 7

from cos x=0,
we have our first solutions:
namely:
\[x=\frac{ \pi }{ 2 }+k \pi\]
where
\[k=0, \pm1, \pm2,\pm3....\]

Answer 8

after that, I don't know!

Answer 9

In This Part 30cosx[1-(tanx)^2] shouldnt u take [cosx]^2 as common factor not cosx

Answer 10

He did, but he factored it out of the whole equation rather than just that portion.

Answer 11

we can try to use the subsequent identity:
\[\frac{ 1 }{ \cos x }=\sqrt{(\tan x)^{2}+1}\]

Answer 12

so our second equation, can be rewritten as below:
\[7 \tan x-20+30\frac{ 1-(\tan x)^{2} }{ \sqrt{1+(\tan x)^{2}} }=0\]
and solving for tan x

Answer 13

Im Inclined to take the first equation answer , but what would be the value for tanx ?

Answer 14

@AmrAhmed please set tan x=z in my second equation, and you will get an irrational equation in z

Answer 15

here is your equation in z:
\[(7z-20)\sqrt{1+z ^{2}}=30z ^{2}-30\]

Answer 16

Well it still doesn't solve it since it gives me areas irregularities in the diagram since x is an intersection point ,good job however .

Answer 17

after squared both sides of the equation above I will get:
\[851 z ^{4}+280 z ^{3}-2249 z ^{2}+280 z +500=0\]
which can be soved using tangents method of Newton, for example

Answer 18

thank you! for your appreciation @AmrAhmed

Answer 19

thank you too @Jhannybean

Answer 20

Np:)