Solve the trig equations 2sin^2(x) + sin(x) - 1 =0

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Homework Statement



Find all values of x in the interval 0<=x<=2pi for which 2sin^2(x)+sin(x)-1=0.

Homework Equations





The Attempt at a Solution



I have no idea.

I spent awhile trying to figure it out on my graphics calculator but couldn't figure it out.

I have only been told how to solve trig equations where there is 1 trig function, lol.

any help would be great thanks.
 
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This equation is quadratic, the variable is sin(x). Can you solve now?
 
ax^2+bx+c=0

quadratic formula ...

for trig

x=trig identity
 
o rite i see.

I think I have the right answers lol

cheeers
 
Trail_Builder said:
o rite i see.

I think I have the right answers lol

cheeers
You can always check it by plugging it back into your original equation.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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