Solve the equation on the interval [0,360) 2sin^2(x)+3sin(x)+1=0

Question

As well as this one below. I'm just having trouble understanding the process and steps. 

 

 Find all the solutions in [0,2pi)
sin^2(t)-cos^2(t)=0

3. Solve the equation on the interval [0,360)

    2sin^2(x)+3sin(x)+1=0

 

4. Find all the solutions in [0,2pi)

    sin^2(t)-cos^2(t)=0

 

5. consider the equation cos(3.14t)=-0.7. Find the smallest positive solution in radians and round your answer to 4 decimal places 

 

6. Consider the equation tan(3.25t)-6.6. Find the smallest positive solution in radians and round your answer to 4 decimal places. 

Answer 1

sin^2(t)-cos^2(t)=0

(sin[t]+cos[t])(sin[t]-cos[t])=0

sin[t]+cos[t]=0

sin(t)=-cos(t)

look on the unit circle

t=3∏/4 and t=7∏/4 or t=135º and t=315º

sin(t)-cos(t)=0

sin(t)=cos(t)

look on the unit circle

t=∏/4 and t=5∏/4 or t=45º and t=225º

 

2sin^2(x)+3sin(x)+1=0

factor

(2sinx+1)(sinx+1)=0

2sinx+1=0

2sinx=-1

sinx=-1/2

look on the unit circle or use arcsin calculator

x=7∏/6 and t=11∏/6 or x=210º and x=330º

sinx+1=0

sinx=-1

look on the unit circle

x=270º

 

5) was done for you; try 6)