For the first one, rewrite the equation using the double-angle identity for sine, to get:
$\displaystyle 2\sin(x)\cos(x)=\sqrt{2}\cos(x)$
Now, divide through by $\displaystyle \sqrt{2}$ then subtract $\displaystyle \cos(x)$ from both sides, factor and use the zero-factor property to equate both factors to zero and solve.
For the second one, use a Pythagorean identity to get a quadratic in $\displaystyle \sin(x)$ :
$\displaystyle 2(1-\sin^2(x))+3\sin(x)=3$
Now, distribute, and arrange in standard quadratic form, and you will find it factors nicely, and use the zero-factor property to equate both factors to zero and solve.
James this is very easy
for the first remember that sin(2x) =2sinxcosx
therefore 2sinxcosx=sqrt(2)cosx
transfer to the frst term and factorize to find
cosx =0 and sinx=sqrt(2)/2 solve them .....
for the second substitute cos^2(x) =1-sin^2(x) and solve the quadratic equation that you will obtain for sinx ..it is easy