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put cos(theta) = x in the first equation and solve for x, and then substitute it back into cos theta
for the second one 2-2sin^theta = 2(1-sin^2theta) = 2 cos^2theta and then put cos theta = x and solve for x
i'm sorry but i don't understand any of that...
substitute cos theta = x in the first problem and post what you got.
as in.. 6x^2+x-1?? like a quadratic?
so solve that quadratic equation.
i dont remember how to when the x^2 value has a coefficient..
Find out and solve it. If you have done it before, I am sure you can solve it.
well everything becomes a fraction if i factor out the 6.... that vid didnt help.
6x2+x-1 = 6x2+3x-2x-1
alright, so now what is the next step?
do you know how to factor quadratic equations?
using the qudratic frmula? yes. why did i need to break up that middle term?
okay. there is another way to solve quadratic eqns. its called factoring. so 6x2+3x-2x-1 = 3x(2x+1)-1(2x+1) = (3x-1)(2x+1)=0. so x = 1/3 or x = -1/2.
ok, but how does this relate to the trig question? haha
so if x = 1/3, it means cos theta = 1/3. and x= -1/2. so cos theta = -1/2. solve foe theta within 0 and 360 degrees
wwhat do you mean by solving for theta between 0 and 360? sorry i'm so bad at this!
just find theta.
dont worry about the 0 and 360.
post the values of theta you got.
arent the x= values the solutions?
No, x is just a variable you used to denote cos theta. so you have to find theta