89.7k views
3 votes
Find all solutions of the equation in the interval [0,2pi).
4cos^(2) x=5-4sinx

1 Answer

1 vote

Remark Change all the trig functions to sin's

Solution

4(1 - sin^2(x)) = 5 - 4sin(x) Remove the brackets.

4 - 4*sin^2(x) = 5 - 4sin(x) Let y = sin(x)

4 - 4y^2 = 5 - 4y Add 4y^2 to both sides; Subtract 4 from both sides.

0 = 5 -4 - 4y + 4y^2

0 = 1 - 4y + 4y^2 'factor

0 = (1 - 2y)^2 Take the square root of both sides.

sqrt(1 - 2y)^2 = sqrt(0)

1 - 2y = 0 Subtract 1 from both sides

-2y = - 1 Divide by - 2

y = -1/-2

y = 1/2 Let sin(x) = y

Sin(x) = 1/2 There are 2 places where sin(x) = 1/2

x = sin-1(1/2)

x = 30o

x = 180 - 30 = 150

Note: the sine function is positive in quadrants 1 and 2. In quadrant 2, the sine function is found by 180 - x. In this case since x = 30, then the other angle giving sin(X) = 1/2 is 180 - 30 = 150

Answer

x = 30

x = 150

User Raja Rao
by
5.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.