Question #21463

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Answer 1 · 2016-11-30T03:11:49

For $0<=theta<2pi$ , $theta = pi/5, (5pi)/6, (3pi)/2$

$cos^2theta-sin^2theta=sintheta$

Use the trig identity $sin^2theta + cos^2theta =1$
$=> cos^2theta=1-sin^2theta$

Substitute for $cos^2theta$

$1-sin^2theta-sin^2theta=sintheta$

$-2sin^2theta +1=sintheta$

$0=2sin^2theta+sintheta-1$

Factor

$(2sintheta-1)(sintheta+1)=0$

Set each factor equal to zero and solve.

$2sintheta - 1 =0 color(white)(aaa)sintheta+1=0$

$sintheta=1/2 color(white)(aaa)sintheta=-1$

For $0<=theta<2pi$ , use unit circle to find

$theta =pi/6, (5pi)/6color(white)(aaa)theta =(3pi)/2$

Answer 2 · 2016-11-30T05:46:09

$cos^2theta-sin^2theta=sintheta$

$=>cos2theta=cos(pi/2-theta)$

$2theta=2npi±(pi/2-theta)," where "n in ZZ$

When $2theta=2npi+(pi/2-theta)$

$=>3theta=(4n+1)pi/2$

$=>theta=(4n+1)pi/6$

Again when

$2theta=2npi-(pi/2-theta)$

$=>2theta=2npi-pi/2+theta$

$=>theta=(4n-1)pi/2$