How to Solve Trigonmetric Equation

**zerobladex** · Jan 30th 2010, 03:14 PM

Hi, I have two equations each which two variables:

x*cos (30) - 275*cos(a) = 0

x*(sin 30) + 275*sin(a) - 300 = 0

I tried solving for the unknowns but eventually I end up with this equation and get stuck.

158.77*cos(a) + 275*sin(a) - 300 = 0

Can someone help me with this? I'm not very good at trigonometric identities...

**VonNemo19** · Jan 30th 2010, 03:37 PM

Originally Posted by zerobladex

Hi, I have two equations each which two variables:

x*cos (30) - 275*cos(a) = 0

x*(sin 30) + 275*sin(a) - 300 = 0

I tried solving for the unknowns but eventually I end up with this equation and get stuck.

158.77*cos(a) + 275*sin(a) - 300 = 0

Can someone help me with this? I'm not very good at trigonometric identities...

Note: $\cos30^\circ=\sqrt{3}{2}$ and $\sin30^\circ=\frac{1}{2}$

Solving for $x$ in the first equation gives $x=\frac{275\cos{a}}{\sqrt3/2}$

Substituting into the second equation

$\left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}+275\sin{a}=0$

Now, we want everything in terms of one trig function, so we use $\sin{a}=\pm\sqrt{1-\cos^2a}$ , then

$\left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}+275(\pm\sqrt{1-\cos^2a})=0$

$\left[\left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}\right]^2=\left(275(\pm\sqrt{1-\cos^2a})\right)^2$

can you finish?

**zerobladex** · Jan 30th 2010, 04:05 PM

I'm still not sure how to solve it, since in the original equation there was a 300 too.

**VonNemo19** · Jan 30th 2010, 04:09 PM

Originally Posted by zerobladex

I'm still not sure how to solve it, since in the original equation there was a 300 too.

Sorry, I mssed that. OK, then:

$\left[\left(\frac{275\cos{a}}{\sqrt3/2}\right)\frac{1}{2}{\color{blue}-300}\right]^2=\left(275(\pm\sqrt{1-\cos^2a})\right)^2$

How bout now?

And it may help if you let $u=\cos{a}$ , such that

$\left[\left(\frac{275{\color{red}u}}{\sqrt3/2}\right)\frac{1}{2}-300\right]^2=\left(275(\pm\sqrt{1-{\color{red}u}^2})\right)^2$

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