Solve the equation cos^2x-sin^2x=sinx+1 in the interval (0<=x<=2pi)?

2 Answers
Oct 4, 2017

x={0,(5pi)/6,pi,(7pi)/6,2pi}

Explanation:

As cos^2x-sin^2x=sinx+1

we have 1-sin^2x-sin^2x=sinx+1

or 2sin^2x+sinx=0

or sinx(2sinx+1)=0

i.e. either sinx=0=sin0 i.e. x=npi

or sinx=-1/2=sin(-pi/6) i.e. x=npi+(-1)^n(-pi/6)

and with in period [0,2pi], we have

x={0,(5pi)/6,pi,(7pi)/6,2pi}

Oct 4, 2017

x=0, (5pi)/6, pi, (7pi)/6, 2pi (0<=x<=2pi)

Explanation:

cos^2x-sin^2x=sinx+1
(1-sin^2x)-sin^2x=sinx+1
-2sin^2x cancel(+1)=sinx cancel(+1)
2sin^2x+sinx=0
sinx(2sinx+1)=0

Here we the above equation to get two new equations (I'm assuming you know how to solve quadratic equations):

sinx=0
therefore x=0, pi, 2pi (0<=x<=2pi)

OR

2sinx+1=0
sinx=-1/2

Here, we cannot go straight to saying x=pi/6, as this would give a positive answer. Sine is negative in the 3rd and 4th quadrants, so these of x in sinx=-1/2 can be found by putting the acute positive angle x=pi/6 into pi+x, and 2pi-x.

So,

2sinx+1=0
sinx=-1/2
x= pi+pi/6, 2pi-pi/6
therefore x=(5pi)/6, (7pi)/6 (0<=x<=2pi)

Putting these two sets of answers together:
therefore x=0, (5pi)/6, pi, (7pi)/6, 2pi (0<=x<=2pi)

However this is assuming that the domain the question asked for is inclusive of 0 and 2pi.

If it is exclusive of these, the solutions to this equation will be:
x= (5pi)/6, pi, (7pi)/6 (0 < x < 2pi)

Solving this is just like solving an algebraic equation once you get everything in terms of sinx, so be sure to know your algebra and trig identities back to front!