Answer:
x = 0
Step-by-step explanation:
You want the solutions on [0, 2π) to ...
[tex]\sin\left(\dfrac{3\pi}{2}+x\right)+\sin\left(\dfrac{3\pi}{2}+x\right)=-2[/tex]
Steps
Step 1 is look at the problem. Here, you notice the arguments of the sine functions are identical, so the two sine function expressions can be combined:
[tex]2\sin\left(\dfrac{3\pi}{2}+x\right)=-2[/tex]
Now, you can divide by 2, with the result that you're looking for an angle whose sine is -1.
[tex]\sin\left(\dfrac{3\pi}{2}+x\right)=-1\\\\\dfrac{3\pi}{2}+x=\sin^{-1}(-1)[/tex]
The sine function has this value only at 3π/2 in the given interval. Once you find that value, you have a one-step linear equation that you can solve in the usual way.
[tex]\dfrac{3\pi}{2}+x=\dfrac{3\pi}{2}\\\\\boxed{x=0}\qquad\text{subtract $\dfrac{3\pi}{2}$}[/tex]
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Additional comment
In general, you solve trig equations of this sort by rewriting them into a form that lets you use an inverse trig function to find the desired angle(s). Sometimes, trig identities are involved.
Here, one could expand the sine of the sum of angles using a trig identity. This would rewrite the equation to 2(sin(3π/2)cos(x) +cos(3π/2)sin(x)) = -2, which simplifies to -cos(x) = -1, or x=0 as above.
