Really need help with how to figure these out.

Attachment 28428

Thanks

James

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- May 22nd 2013, 03:46 AMJamesTSolutions for sin and cos equations
Really need help with how to figure these out.

Attachment 28428

Thanks

James - May 22nd 2013, 03:57 AMMarkFLRe: Solutions for sin and cos equations
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. - May 22nd 2013, 03:57 AMMINOANMANRe: Solutions for sin and cos equations
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 - May 22nd 2013, 04:03 AMJamesTRe: Solutions for sin and cos equations
omg, thank you so much Mark and Mino!!!